hdu 2389 Rain on your Parade【最大匹配】

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题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=2389
用 Hopcroft-Karp 算法
匈牙利算法会超时
代码：
#include <stdio.h>#include <iostream>#include <math.h>#include <stdlib.h>#include <ctype.h>#include <algorithm>#include <vector>#include <string.h>#include <string>#include <queue>#include <stack>#include <set>#include <sstream>#include <time.h>using namespace std;const int N = 3005;const int INF = 1 << 28;int g[N][N];int Mx[N];int My[N];int dx[N];int dy[N];bool used[N];int Nx, Ny, dis;bool searchP(){    dis = INF;    int i, v, u;    std::queue<int> Q;    memset(dx, -1, sizeof(dx));    memset(dy, -1, sizeof(dy));    for (i = 0; i<Nx; i++)    {        if (Mx[i] == -1)        {            Q.push(i);            dx[i] = 0;        }    }    while (!Q.empty())    {        u = Q.front();        Q.pop();        if (dx[u]>dis) break;        for (v = 0; v<Ny; v++)        {            if (g[u][v] && dy[v] == -1)            {                dy[v] = dx[u] + 1;                if (My[v] == -1) dis = dy[v];                else                {                    dx[My[v]] = dy[v] + 1;                    Q.push(My[v]);                }            }        }    }    return dis != INF;}bool DFS(int u){    int v;    for (v = 0; v<Ny; v++)    {        if (g[u][v] && !used[v] && dy[v] == dx[u] + 1)        {            used[v] = true;            if (My[v] != -1 && dy[v] == dis) continue;            if (My[v] == -1 || DFS(My[v]))            {                My[v] = u;                Mx[u] = v;                return true;            }        }    }    return false;}int Hungary(){    int u;    int ret = 0;    memset(Mx, -1, sizeof(Mx));    memset(My, -1, sizeof(My));    while (searchP())    {        memset(used, false, sizeof(used));        for (u = 0; u<Nx; u++)            if (Mx[u] == -1 && DFS(u))  ret++;    }    return ret;}struct Peo{    int x;    int y;    int s;}peo[3010];int n, m;int main(){    int t, cases = 1;    scanf("%d", &t);    while (t--)    {        memset(g, 0, sizeof(g));        int T, x, y;        scanf("%d", &T);        scanf("%d", &m);        for (int i = 1;i <= m;i++)            scanf("%d%d%d", &peo[i].x, &peo[i].y, &peo[i].s);        scanf("%d", &n);        for (int i = 1;i <= n;i++)        {            scanf("%d%d", &x, &y);            for (int j = 1;j <= m;j++)            {                int Dis = ceil(sqrt((peo[j].x - x) * (peo[j].x - x) + (peo[j].y - y) * (peo[j].y - y)) * 1.0 / peo[j].s);                if (Dis <= T)                    g[i-1][j-1] = 1;            }        }        Nx = n;        Ny = m;        int ans = Hungary();        printf("Scenario #%d:\n%d\n\n", cases++, ans);    }    return 0;}
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