POJ 2536 Gopher II(二分图最大匹配)

题目地址
题意：n只地鼠，m个洞，老鹰的到达地面的时间s，地鼠的移动速度v，求多少只地鼠会被老鹰吃了。
思路：直接暴力枚举每个地鼠到哪些洞不会被老鹰吃掉，因为每个洞只能有一只地鼠，所以求出的最大匹配就是不被老鹰吃掉的地鼠个数，然后用n减去就好了。

#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <stack>#include <cmath>#include <cstdio>#include <algorithm>#define N 2010#define LL __int64#define inf 0x3f3f3f3f#define lson l,mid,ans<<1#define rson mid+1,r,ans<<1|1#define getMid (l+r)>>1#define movel ans<<1#define mover ans<<1|1using namespace std;const LL mod = 1e9 + 7;int n, m;struct node{    double x, y;}gopher[N], gopher_holes[N];struct Hungarian {    vector<int> mapp[N];    bool vis[N];//是否为匹配点    int mark[N];//该边与哪个点构成的边为匹配边    void init() {        for (int i = 0; i <= n + m + 20; i++) {            mapp[i].clear();        }        memset(mark, -1, sizeof(mark));    }    void add(int a, int b) {        mapp[a].push_back(b);        mapp[b].push_back(a);    }    bool dfs(int u) {        for (int i = 0; i < mapp[u].size(); i++) {            int v = mapp[u][i];            if (!vis[v]) {                vis[v] = true;                if (mark[v] == -1 || dfs(mark[v])) {                    mark[u] = v;                    mark[v] = u;                    return true;                }            }        }        return false;    }    int solve() {        int sum = 0;        for (int i = 1; i <= n; i++) {//枚举非匹配点            if (mark[i] == -1) {                memset(vis, false, sizeof(vis));                if (dfs(i)) {                    sum++;//最大匹配的个数                }            }        }        return sum;    }}hungarian;double dis(node p1, node p2){    return  sqrt((p2.x - p1.x)*(p2.x - p1.x) + (p2.y - p1.y)*(p2.y - p1.y));}int main() {    int a, b, c;    int s, v;    cin.sync_with_stdio(false);    while (cin >> n >> m >> s >> v) {        hungarian.init();        for (int i = 1; i <= n; i++) {            cin >> gopher[i].x >> gopher[i].y;        }        for (int i = 1; i <= m; i++) {            cin >> gopher_holes[i].x >> gopher_holes[i].y;        }        for (int i = 1; i <= n; i++) {            for (int j = 1; j <= m; j++) {                double lens = dis(gopher_holes[j], gopher[i]);                if (lens / v <= (double)s) {                    hungarian.add(i, n + j);                }            }        }        cout << n - hungarian.solve() << endl;    }    return 0;}