hdoj 3622 Bomb Game

類型：2-sat

題目：http://acm.hdu.edu.cn/showproblem.php?pid=3622

來源：2010 Asia Regional Tianjin Site —— Online Contest

思路：二分枚舉半徑，判斷解是否存在

// hdoj 3622 Bomb Game
// wa wa 1984MS	612K
#include <iostream>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define FOR(i,a,b) for(i = (a); i < (b); ++i)
#define FORE(i,a,b) for(i = (a); i <= (b); ++i)
#define FORD(i,a,b) for(i = (a); i > (b); --i)
#define FORDE(i,a,b) for(i = (a); i >= (b); --i)
#define CLR(a,b) memset(a,b,sizeof(a))
#define PB(x) push_back(x)

const int N = 501;
const int M = 40010;
const int INF = 0x7f7f7f7f;
const double eps = 1e-8;

int low[N], dfn[N];
int belong[N];
bool inStack[N], vis[N];
stack<int> st;
int n, m, step, t;
int cnt1;
int head1[N], p[N][4];
double max_dis;
struct edge1 {
    int v, nxt;
}E1[M];

void addedge1(int u, int v) {
    E1[cnt1].v = v;
    E1[cnt1].nxt = head1[u];
    head1[u] = cnt1++;
}

double dis(double x1, double y1, double x2, double y2) {
    return sqrt((x1 - x2) * (x1 - x2) + (y1- y2) * (y1 - y2));
}

void tarjan(int u) {
    int i;
    step++;
    st.push(u);
    low[u] = dfn[u] = step;
    vis[u] = 1;
    inStack[u] = 1;
    for(i = head1[u]; i != -1; i = E1[i].nxt) {
        int x = E1[i].v;
        if(!vis[x]) {
            tarjan(x);
            low[u] = min(low[u], low[x]);
        }
        else if(inStack[x])
            low[u]=min(low[u], dfn[x]);
    }
    if(low[u] == dfn[u]) {
        t++;
        while(1) {
            int x = st.top();
            st.pop();
            belong[x] = t;
            inStack[x] = 0;
            if(x == u)
                break;
        }
    }
}

bool is_ok(double dis, double x) {
    if(x / 2.0 - dis >= eps)
        return true;
    return false;
}

void solve() {
    int i, j;
    double mid, l = 0, r = max_dis;

    while(r - l > eps) {
        mid = (l + r) / 2.0;
        cnt1 = step = t = 0;
        CLR(head1, -1);
        CLR(vis, 0);
        CLR(inStack, 0);
        CLR(belong, 0);
        FORE(i, 1, n) FORE(j, i + 1, n) {
            if(!is_ok(mid, dis(p[i][0], p[i][1], p[j][0], p[j][1]))) {
                addedge1(i, j + n);
                addedge1(j, i + n);
            }
            if(!is_ok(mid, dis(p[i][0], p[i][1], p[j][2], p[j][3]))) {
                addedge1(i, j);
                addedge1(j + n, i + n);
            }
            if(!is_ok(mid, dis(p[i][2], p[i][3], p[j][0], p[j][1]))) {
                addedge1(i + n, j + n);
                addedge1(j, i);
            }
            if(!is_ok(mid, dis(p[i][2], p[i][3], p[j][2], p[j][3]))) {
                addedge1(i + n, j);
                addedge1(j + n, i);
            }
        }
        while(!st.empty())
            st.pop();
        FORE(i, 1, 2 * n)
            if(!vis[i])
                tarjan(i);
        bool sign = false;
        FORE(i, 1, n)
            if(belong[i] == belong[i + n]) {
                sign = true;
                break;
            }
        if(sign)
            r = mid;
        else
            l = mid;
    }
    printf("%.2lf\n", l);
}

int main() {
    int u, v, i, j;

    while(scanf("%d", &n) == 1) {
        max_dis = -INF;
        FORE(i, 1, n) {
            scanf("%d %d %d %d", &p[i][0], &p[i][1], &p[i][2], &p[i][3]);
            max_dis = max(max_dis, dis(p[i][0], p[i][1], p[i][2], p[i][3]));
        }
        solve();
    }
    return 0;
}