hdu4215 Moles

不錯的題目

笛卡爾樹+DFS+KMP匹配

這道題建樹複雜度過高，所以需要笛卡爾樹（隊友敲的這部分，不懂）

然後dfs得到模式串再直接匹配就好了

#pragma comment(linker, "/STACK:1024000000,1024000000")
//HEAD
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
//LOOP
#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
//STL
#define SZ(V) (int)V.size()
#define PB push_back
#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)
#define ALL(c) (c).begin(), (c).end()
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RS(s) scanf("%s", s)
//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)
typedef long long LL;
typedef unsigned long long ULL;
typedef vector <int> VI;
const int INF = 100000000;
const double eps = 1e-10;
const int maxn = 1805000;
const LL MOD = 1e9 + 7;


const int MAXN = 600010;

struct data{
    int k, a, num;
}d[MAXN];
int val[MAXN];
bool cmp(data x, data y)
{
    return x.k < y.k;
}
int convert[MAXN];
struct Tree{
    int fa;
    int lson, rson;
}t[MAXN];

int n, cnt;
string ans;
int f[maxn];

void _insert(int i)
{
    int j = d[i - 1].num;
    while (j && d[convert[j]].a > d[i].a) { j = t[j].fa; }

    int p = d[i].num;
    int b = t[j].rson;///初始0 的rson 爲0；
    t[j].rson = p;
    t[p].fa = j;
//    if (b)///必須
//    {
        t[b].fa = p;
        t[p].lson = b;
//    }
}

void dfs(int x)
{
    ans += ((val[x] & 1) ? '1' : '0');
    if (t[x].lson != -1)
    {
        dfs(t[x].lson);
        ans += ((val[x] & 1) ? '1' : '0');
    }
    if (t[x].rson != -1)
    {
        dfs(t[x].rson);
        ans += ((val[x] & 1) ? '1' : '0');
    }
}

void getFail(string P, int *f)
{
    int m = P.size();
    f[0] = 0;
    f[1] = 0;
    for (int i = 1; i < m; i++)
    {
        int j = f[i];
        while (j && P[i] != P[j]) j = f[j];
        f[i + 1] = P[i] == P[j] ? j + 1 : 0;
    }
}

int find(string T, string P, int *f)
{
    int ret = 0;
    int n = T.size();
    int m = P.size();
    getFail(P, f);
    int j = 0;
    for (int i = 0; i < n; i++)
    {
        while (j && P[j] != T[i]) j = f[j];
        if (P[j] == T[i]) j++;
        if (j == m) ret++;
    }
    return ret;
}
int main ()
{
    int T;
    int ncase = 1;
    RI(T);
    while(T--)
    {
        RI(n);
//        CLR(t, 0);
        CLR(t, -1);///-1
        FE(i, 1, n)
        {
            RI(d[i].k);
            val[i] = d[i].k;
            d[i].a = i;
            d[i].num = i;
        }
        sort(d + 1, d + 1 + n, cmp);
        ///默認點0
        FE(i, 1, n) convert[d[i].num] = i;
        FE(i, 1, n) _insert(i);
        t[1].fa = -1;
        ans.clear(), cnt = 1;
        dfs(1);
        string ca;
        cin >> ca;
        printf("Case #%d: %d\n", ncase++, find(ans, ca, f));
    }
    return 0;
}