舞蹈鏈(DLX)

算法講解

在kuangbin第三套裏面學習了DLX。
這個東西說明白了就是精確覆蓋和重複覆蓋的特殊搜索剪枝。所以複雜度貼近暴搜，比賽裏面我還沒見過用DLX的題目。
具體DLX的原理我這就不寫了，給大家推薦個寫的好的大佬博客~
個人覺得DLX幾乎可以當作黑盒用。只要轉化成行座標是你的所有選擇，列座標所有是你需要覆蓋的，每行是每個選擇可以覆蓋的，就ok了。
這裏先給大家貼下我的精確覆蓋和重複覆蓋的模板

精確覆蓋

const int MN = 1005;//最大行數
const int MM = 1005;//最大列數
const int MNN = (int) 1e5 + 5 + MM; //最大點數

struct DLX {
    int n, m, si;
    int U[MNN], D[MNN], L[MNN], R[MNN], Row[MNN], Col[MNN];
    int H[MN], S[MM];
    int ansd, ans[MN];

    void init(int _n, int _m) {
        n = _n;
        m = _m;
        for (int i = 0; i <= m; i++) {
            S[i] = 0;
            U[i] = D[i] = i;
            L[i] = i - 1;
            R[i] = i + 1;
        }
        R[m] = 0;
        L[0] = m;
        si = m;
        for (int i = 1; i <= n; i++)
            H[i] = -1;
    }

    void link(int r, int c) {

        ++S[Col[++si] = c];
        Row[si] = r;
        D[si] = D[c];
        U[D[c]] = si;
        U[si] = c;
        D[c] = si;
        if (H[r] < 0)
            H[r] = L[si] = R[si] = si;
        else {
            R[si] = R[H[r]];
            L[R[H[r]]] = si;
            L[si] = H[r];
            R[H[r]] = si;
        }
    }

    void remove(int c) {

        L[R[c]] = L[c];
        R[L[c]] = R[c];
        for (int i = D[c]; i != c; i = D[i])
            for (int j = R[i]; j != i; j = R[j]) {
                U[D[j]] = U[j];
                D[U[j]] = D[j];
                --S[Col[j]];
            }
    }

    void resume(int c) {

        for (int i = U[c]; i != c; i = U[i])
            for (int j = L[i]; j != i; j = L[j])
                ++S[Col[U[D[j]] = D[U[j]] = j]];
        L[R[c]] = R[L[c]] = c;
    }

    bool dance(int d) {

        if (ansd != -1 && ansd < d)return 0;
        if (R[0] == 0) {

            if (ansd == -1)ansd = d;
            else if (d < ansd) ansd = d;
            return 1;
        }
        int c = R[0];
        for (int i = R[0]; i != 0; i = R[i])
            if (S[i] < S[c])
                c = i;
        remove(c);
        for (int i = D[c]; i != c; i = D[i]) {
            ans[d] = Row[i];
            for (int j = R[i]; j != i; j = R[j])
                remove(Col[j]);
            (dance(d + 1));
            for (int j = L[i]; j != i; j = L[j])
                resume(Col[j]);
        }
        resume(c);
        return 0;
    }
} dlx;

重複覆蓋

const int MN = 55;//最大行數
const int MM = 55;//最大列數
const int MNN = MM * MN + MM + MN + 10; //最大點數

struct DLX {
    int n, m, si;
    int U[MNN], D[MNN], L[MNN], R[MNN], Row[MNN], Col[MNN];
    int H[MN], S[MM];
    int ansd, ans[MN];

    void init(int _n, int _m) {
        n = _n;
        m = _m;
        for (int i = 0; i <= m; i++) {
            S[i] = 0;
            U[i] = D[i] = i;
            L[i] = i - 1;
            R[i] = i + 1;
        }
        R[m] = 0;
        L[0] = m;
        si = m;
        for (int i = 1; i <= n; i++)
            H[i] = -1;
    }

    void link(int r, int c) {
        ++S[Col[++si] = c];
        Row[si] = r;
        D[si] = D[c];
        U[D[c]] = si;
        U[si] = c;
        D[c] = si;
        if (H[r] < 0)
            H[r] = L[si] = R[si] = si;
        else {
            R[si] = R[H[r]];
            L[R[H[r]]] = si;
            L[si] = H[r];
            R[H[r]] = si;
        }
    }

    void remove(int c) {
        for (int i = D[c]; i != c; i = D[i]) {
            L[R[i]] = L[i], R[L[i]] = R[i];
        }
    }

    void resume(int c) {
        for (int i = U[c]; i != c; i = U[i]) {
            L[R[i]] = R[L[i]] = i;
        }
    }

    int f_check() {
        int vis[MNN];
        memset(vis, 0, sizeof(vis));
        int ret = 0;
        for (int c = R[0]; c != 0; c = R[c]) vis[c] = true;
        for (int c = R[0]; c != 0; c = R[c])
            if (vis[c]) {
                ret++;
                vis[c] = false;
                for (int i = D[c]; i != c; i = D[i])
                    for (int j = R[i]; j != i; j = R[j])
                        vis[Col[j]] = false;
            }
        return ret;
    }

    bool dance(int d, int limit) {
        if (d > limit)return false;
        if (d + f_check() > limit)return false;
        if (R[0] == 0) {
            ansd = d;
            return true;
        }
        int c = R[0];
        for (int i = R[0]; i != 0; i = R[i])
            if (S[i] < S[c])
                c = i;
        for (int i = D[c]; i != c; i = D[i]) {
            ans[d] = Row[i];
            remove(i);
            for (int j = R[i]; j != i; j = R[j])
                remove(j);
            if (dance(d + 1, limit)) return true;
            for (int j = L[i]; j != i; j = L[j])
                resume(j);
            resume(i);
        }
        return false;
    }
} dlx;

題解

我這裏面主要就把kuangbin套題前幾題的題解稍微寫寫。

ZOJ-3209 Treasure Map

傳送門：ZOJ-3209

題意

給你一個n*m的矩陣，再給你p個小矩陣，告訴你每個矩陣的左下角座標和右上角座標，要求你用小矩陣覆蓋大矩陣，同時小矩陣不能重疊，求至少用幾個矩陣。

題解

把小矩陣的編號作爲行，把所有點作爲列，這樣就轉化成了精確覆蓋問題了，模板一通xjb套，ac～

ac代碼

#include <bits/stdc++.h>

using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = (int) 1e6 + 7;
typedef long long ll;

const int MN = 1005;//最大行數
const int MM = 1005;//最大列數
const int MNN = (int) 1e5 + 5 + MM; //最大點數

struct DLX {
    int n, m, si;
    int U[MNN], D[MNN], L[MNN], R[MNN], Row[MNN], Col[MNN];
    int H[MN], S[MM];
    int ansd, ans[MN];

    void init(int _n, int _m) {
        n = _n;
        m = _m;
        for (int i = 0; i <= m; i++) {
            S[i] = 0;
            U[i] = D[i] = i;
            L[i] = i - 1;
            R[i] = i + 1;
        }
        R[m] = 0;
        L[0] = m;
        si = m;
        for (int i = 1; i <= n; i++)
            H[i] = -1;
    }

    void link(int r, int c) {

        ++S[Col[++si] = c];
        Row[si] = r;
        D[si] = D[c];
        U[D[c]] = si;
        U[si] = c;
        D[c] = si;
        if (H[r] < 0)
            H[r] = L[si] = R[si] = si;
        else {
            R[si] = R[H[r]];
            L[R[H[r]]] = si;
            L[si] = H[r];
            R[H[r]] = si;
        }
    }

    void remove(int c) {

        L[R[c]] = L[c];
        R[L[c]] = R[c];
        for (int i = D[c]; i != c; i = D[i])
            for (int j = R[i]; j != i; j = R[j]) {
                U[D[j]] = U[j];
                D[U[j]] = D[j];
                --S[Col[j]];
            }
    }

    void resume(int c) {

        for (int i = U[c]; i != c; i = U[i])
            for (int j = L[i]; j != i; j = L[j])
                ++S[Col[U[D[j]] = D[U[j]] = j]];
        L[R[c]] = R[L[c]] = c;
    }

    bool dance(int d) {

        if (ansd != -1 && ansd < d)return 0;
        if (R[0] == 0) {

            if (ansd == -1)ansd = d;
            else if (d < ansd) ansd = d;
            return 1;
        }
        int c = R[0];
        for (int i = R[0]; i != 0; i = R[i])
            if (S[i] < S[c])
                c = i;
        remove(c);
        for (int i = D[c]; i != c; i = D[i]) {
            ans[d] = Row[i];
            for (int j = R[i]; j != i; j = R[j])
                remove(Col[j]);
            (dance(d + 1));
            for (int j = L[i]; j != i; j = L[j])
                resume(Col[j]);
        }
        resume(c);
        return 0;
    }
} dlx;


int main() {

#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif

    int T;
    cin >> T;
    while (T--) {
        int n, m, p;
        cin >> n >> m >> p;
        dlx.init(p, n * m);
        for (int i = 1; i <= p; i++) {
            int x1, y1, x2, y2;
            cin >> x1 >> y1 >> x2 >> y2;
            for (int j = x1 + 1; j <= x2; j++) {
                for (int k = y1 + 1; k <= y2; k++) {
                    dlx.link(i, (j - 1) * m + k);
                }
            }
        }
        dlx.ansd = -1;
        dlx.dance(0);
        cout << dlx.ansd << endl;
    }

    return 0;
}

HDU - 2295 Chef and Graph Queries

傳送門：HDU-2295

題意

給了n個點和m個圓心，在用不多於k個圓的情況下，使得所有的點被覆蓋，最小的圓半徑多大

題解

就二分一下答案，判斷條件就是用了小於等於k個圓，行爲圓，列爲點，重複覆蓋往上一套，注意一下精度，ac～

ac代碼

#include <bits/stdc++.h>

using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = (int) 1e6 + 7;
typedef long long ll;
#define clr(x) memset(x,0,sizeof(x));

struct rec {
    int x, y;
};

rec city[maxn], rad[maxn];
double dis[55][55];

const int MN = 55;//最大行數
const int MM = 55;//最大列數
const int MNN = MM * MN + MM + MN + 10; //最大點數

struct DLX {
    int n, m, si;
    int U[MNN], D[MNN], L[MNN], R[MNN], Row[MNN], Col[MNN];
    int H[MN], S[MM];
    int ansd, ans[MN];

    void init(int _n, int _m) {
        n = _n;
        m = _m;
        for (int i = 0; i <= m; i++) {
            S[i] = 0;
            U[i] = D[i] = i;
            L[i] = i - 1;
            R[i] = i + 1;
        }
        R[m] = 0;
        L[0] = m;
        si = m;
        for (int i = 1; i <= n; i++)
            H[i] = -1;
    }

    void link(int r, int c) {
        ++S[Col[++si] = c];
        Row[si] = r;
        D[si] = D[c];
        U[D[c]] = si;
        U[si] = c;
        D[c] = si;
        if (H[r] < 0)
            H[r] = L[si] = R[si] = si;
        else {
            R[si] = R[H[r]];
            L[R[H[r]]] = si;
            L[si] = H[r];
            R[H[r]] = si;
        }
    }

    void remove(int c) {
        for (int i = D[c]; i != c; i = D[i]) {
            L[R[i]] = L[i], R[L[i]] = R[i];
        }
    }

    void resume(int c) {
        for (int i = U[c]; i != c; i = U[i]) {
            L[R[i]] = R[L[i]] = i;
        }
    }

    int f_check() {
        int vis[MNN];
        memset(vis, 0, sizeof(vis));
        int ret = 0;
        for (int c = R[0]; c != 0; c = R[c]) vis[c] = true;
        for (int c = R[0]; c != 0; c = R[c])
            if (vis[c]) {
                ret++;
                vis[c] = false;
                for (int i = D[c]; i != c; i = D[i])
                    for (int j = R[i]; j != i; j = R[j])
                        vis[Col[j]] = false;
            }
        return ret;
    }

    bool dance(int d, int limit) {
        if (d > limit)return false;
        if (d + f_check() > limit)return false;
        if (R[0] == 0) {
            ansd = d;
            return true;
        }
        int c = R[0];
        for (int i = R[0]; i != 0; i = R[i])
            if (S[i] < S[c])
                c = i;
        for (int i = D[c]; i != c; i = D[i]) {
            ans[d] = Row[i];
            remove(i);
            for (int j = R[i]; j != i; j = R[j])
                remove(j);
            if (dance(d + 1, limit)) return true;
            for (int j = L[i]; j != i; j = L[j])
                resume(j);
            resume(i);
        }
        return false;
    }
} dlx;

int main() {

#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif

    int T, n, m, k;
    cin >> T;
    while (T--) {
        clr(city);
        clr(rad);
        clr(dis);
        double mx = 0;
        cin >> n >> m >> k;
        for (int i = 1; i <= n; i++) {
            cin >> city[i].x >> city[i].y;
        }
        for (int i = 1; i <= m; i++) {
            cin >> rad[i].x >> rad[i].y;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                double x = 1.0 * city[i].x - rad[j].x;
                double y = 1.0 * city[i].y - rad[j].y;
                dis[i][j] = sqrt(x * x + y * y);
                mx = max(dis[i][j], mx);
            }
        }
        double r = mx * 2;
        double l = 0;
        while (r - l > 1e-8) {
            double mid = (r + l) / 2;
            dlx.init(m, n);
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= m; j++) {
                    if (dis[i][j] < mid) dlx.link(j, i);
                }
            }
            dlx.ansd = -1;
            if (dlx.dance(0, k)) r = mid;
            else l = mid;
        }
        printf("%.6lf\n", l);
    }

    return 0;
}

FZU - 1686 神龍的難題

傳送門：FZU - 1686

題意

中文題題我就不寫題意了哈，我相信沒有外國人看我博客～

題解

很經典的一道重複覆蓋～
和第一題有點像吧，就像第一題一樣作，只是別忘了把沒有怪物的點去掉就ok了。去掉的辦法就是構建一個只有內些沒有怪物的點的覆蓋，先把這個覆蓋放進去，就完成構造了。剩下的構造就和第一題完全一樣咯～

ac代碼

#include <cstdio>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <stack>
#include <vector>
#include <iostream>

using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = (int) 1e6 + 7;
typedef long long ll;
#define clr(x) memset(x,0,sizeof(x));

struct rec {
    int x, num;
} mp[20][20];


const int MN = 250;//最大行數
const int MM = 250;//最大列數
const int MNN = MM * MN; //最大點數

struct DLX {
    int n, m, si;
    int U[MNN], D[MNN], L[MNN], R[MNN], Row[MNN], Col[MNN];
    int H[MN], S[MM];
    int ansd, ans[MN];

    void init(int _n, int _m) {
        n = _n;
        m = _m;
        for (int i = 0; i <= m; i++) {
            S[i] = 0;
            U[i] = D[i] = i;
            L[i] = i - 1;
            R[i] = i + 1;
        }
        R[m] = 0;
        L[0] = m;
        si = m;
        for (int i = 1; i <= n; i++)
            H[i] = -1;
    }

    void link(int r, int c) {
        ++S[Col[++si] = c];
        Row[si] = r;
        D[si] = D[c];
        U[D[c]] = si;
        U[si] = c;
        D[c] = si;
        if (H[r] < 0)
            H[r] = L[si] = R[si] = si;
        else {
            R[si] = R[H[r]];
            L[R[H[r]]] = si;
            L[si] = H[r];
            R[H[r]] = si;
        }
    }

    void remove(int c) {
        for (int i = D[c]; i != c; i = D[i]) {
            L[R[i]] = L[i], R[L[i]] = R[i];
        }
    }

    void resume(int c) {
        for (int i = U[c]; i != c; i = U[i]) {
            L[R[i]] = R[L[i]] = i;
        }
    }

    bool vis[MNN];
    int f_check() {
        memset(vis, 0, sizeof(vis));
        int ret = 0;
        for (int c = R[0]; c != 0; c = R[c]) vis[c] = true;
        for (int c = R[0]; c != 0; c = R[c])
            if (vis[c]) {
                ret++;
                vis[c] = false;
                for (int i = D[c]; i != c; i = D[i])
                    for (int j = R[i]; j != i; j = R[j])
                        vis[Col[j]] = false;
            }
        return ret;
    }

    void dance(int d) {
        if (d + f_check() >= ansd)return;
        if (R[0] == 0) {
            if (ansd > d) ansd = d;
            return;
        }
        int c = R[0];
        for (int i = R[0]; i != 0; i = R[i])
            if (S[i] < S[c])
                c = i;
        for (int i = D[c]; i != c; i = D[i]) {
            ans[d] = Row[i];
            remove(i);
            for (int j = R[i]; j != i; j = R[j])
                remove(j);
            dance(d + 1);
            for (int j = L[i]; j != i; j = L[j])
                resume(j);
            resume(i);
        }
    }
} dlx;

int main() {

#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif

    int n, m;
    while (cin >> n >> m) {
        clr(mp);
        int tot = 0;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                int q;
                cin >> q;
                mp[i][j].x = q;
                if (q == 1) {
                    tot++;
                    mp[i][j].num = tot;
                }
            }
        }
        int x, y;
        cin >> x >> y;
        dlx.init(n * m, tot);
        int num = 0;
        for (int i = 1; i <= n - x + 1; i++) {
            for (int j = 1; j <= m - y + 1; j++) {
                num++;
                for (int r = i; r <= i + x - 1; r++) {
                    for (int c = j; c <= j + y - 1; c++) {
                        if (mp[r][c].x) {
                            dlx.link(num, mp[r][c].num);
                        }
                    }
                }
            }
        }

        dlx.ansd = inf;
        dlx.dance(0);
        cout << dlx.ansd << endl;

    }

    return 0;
}

POJ-1084 Square Destroyer

傳送門:POJ-1084

題意

給出一個n*n的大網格，然後它由一系列的火柴拼成，題中給的圖片是第二組測試數據。現在我們已經刪掉了若干個火柴，問至少還需要刪掉多少火柴，才能使圖中的若干個正方形全部被破壞。

題解

這個題的建圖可是不是一般的噁心，每個正方形都需要有其上的至少一根火柴被選，那麼就轉化成了DLX模型。每行是一跟火柴，每列是一個正方形，當然正方形個數是<=5*5+4*4+3*3+2*2+1*1=55的，然後顯然若某行這根火柴在某列的正方形中，則new一個節點。這樣就可以直接套板子就ok了。

ac代碼

#include<iostream>
#include<cstring>

using namespace std;

const int INF = 10e8;
const int MaxN = 70;
const int MaxM = 70;
const int MaxNode = MaxN * MaxM;

struct DLX {
    int L[MaxNode], R[MaxNode], U[MaxNode], D[MaxNode], col[MaxNode], row[MaxNode];
    int S[MaxM], H[MaxN];
    int n, m, size;
    int ans;

    void init(int _n, int _m) {
        n = _n;
        m = _m;

        for (int i = 0; i <= m; ++i) {
            U[i] = D[i] = i;
            R[i] = i + 1;
            L[i] = i - 1;
            row[i] = 0;                // !!!

            S[i] = 0;
        }

        R[m] = 0;
        L[0] = m;

        size = m;
        ans = INF;

        for (int i = 0; i <= n; ++i)        // !!!
            H[i] = -1;
    }

    void Link(int r, int c) {
        col[++size] = c;
        ++S[c];
        row[size] = r;

        U[size] = U[c];
        D[size] = c;
        D[U[c]] = size;
        U[c] = size;

        if (H[r] == -1)
            H[r] = L[size] = R[size] = size;
        else {
            L[size] = L[H[r]];
            R[size] = H[r];
            R[L[H[r]]] = size;
            L[H[r]] = size;
        }
    }

    void remove(int c) {
        for (int i = D[c]; i != c; i = D[i]) {
            R[L[i]] = R[i];
            L[R[i]] = L[i];
        }
    }

    void remove1(int r) {
        if (H[r] == -1)
            return;

        for (int i = U[H[r]]; i != H[r]; i = U[i]) {
            if (H[row[i]] == i)            // !!!
            {
                if (R[i] == i)
                    H[row[i]] = -1;
                else
                    H[row[i]] = R[i];
            }

            L[R[i]] = L[i];
            R[L[i]] = R[i];
        }

        for (int i = R[H[r]]; i != H[r]; i = R[i])
            for (int j = U[i]; j != i; j = U[j]) {
                if (H[row[j]] == j) {
                    if (R[j] == j)
                        H[row[j]] = -1;
                    else
                        H[row[j]] = R[j];
                }

                L[R[j]] = L[j];
                R[L[j]] = R[j];
            }
    }

    void resume(int c) {
        for (int i = U[c]; i != c; i = U[i])
            R[L[i]] = L[R[i]] = i;
    }

    bool vis[MaxM];

    int getH() {
        int ret = 0;

        for (int c = R[0]; c != 0; c = R[c])
            vis[c] = 1;

        for (int c = R[0]; c != 0; c = R[c])
            if (vis[c]) {
                ++ret;
                vis[c] = 0;

                for (int i = D[c]; i != c; i = D[i])
                    for (int j = R[i]; j != i; j = R[j])
                        vis[col[j]] = 0;
            }

        return ret;
    }

    void Dance(int d) {
        if (d + getH() >= ans)
            return;

        if (R[0] == 0) {
            if (d < ans)
                ans = d;

            return;
        }

        int c = R[0];

        for (int i = R[0]; i != 0; i = R[i])
            if (S[i] < S[c])
                c = i;

        for (int i = D[c]; i != c; i = D[i]) {
            remove(i);

            for (int j = R[i]; j != i; j = R[j])
                remove(j);

            Dance(d + 1);

            for (int j = L[i]; j != i; j = L[j])
                resume(j);

            resume(i);
        }
    }
}dlx;

int N;
int ans1[6] = {0, 1, 3, 6, 9, 14};

void slove() {
    dlx.init(2 * N * (N + 1), N * (N + 1) * (2 * N + 1) / 6);

    int t1, t2;
    int cou = 0;
    int K, a;

    cin >> K;

    if (K == 0) {
        cout << ans1[N] << endl;
        return;
    }

    for (int i = 1; i <= N; ++i)
        for (int j = 1; j <= (N - i + 1) * (N - i + 1); ++j) {
            ++cou;
            t1 = (j - 1) % (N - i + 1) + 1 + (2 * N + 1) * ((j - 1) / (N - i + 1));
            for (int k = 0; k < i; ++k) {
                dlx.Link(k + t1, cou);
                dlx.Link((2 * N + 1) * k + t1 + i - 1 + N + 1, cou);
                dlx.Link((2 * N + 1) * k + t1 + N, cou);
                dlx.Link(k + t1 + i * (2 * N + 1), cou);
            }
        }

    for (int i = 0; i < K; ++i) {
        cin >> a;
        dlx.remove1(a);
    }

    dlx.Dance(0);

    if (dlx.ans == INF)
        cout << 0 << endl;
    else
        cout << dlx.ans << endl;
}

int main() {
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    while (T--) {
        cin >> N;

        slove();
    }
    return 0;
}

小計

自己也是真懶，隔了這麼久才寫博客～大家捧場啦～～～

舞蹈鏈

舞蹈鏈(DLX)

算法講解

精確覆蓋

重複覆蓋

題解

ZOJ-3209 Treasure Map

題意

題解

ac代碼

HDU - 2295 Chef and Graph Queries

題意

題解

ac代碼

FZU - 1686 神龍的難題

題意

題解

ac代碼

POJ-1084 Square Destroyer

題意

題解

ac代碼

小計

BZOJ - 1002 輪狀病毒

UVA - 11426 GCD - Extreme (II)

LightOJ - 1245 Harmonic Number (II)(除法分塊)

LightOJ - 1282 Leading and Trailing

HDU - 5478 Can you find it

Mac下配置sublime實現LaTeX

https://yachay.unat.edu.pe/blog/index.php?comment_area=format_blog&comment_component=blog&comment_co

linux以太網驅動總結