HDU 3491 Thieves（经典拆点建图，割点）

题目地址
题意：有一群小偷要从s点逃到t点，告诉你这个城市的道路图，并且告诉你每个关键点要多少警察才不会让小偷逃走，问你最少要多少警察才能阻止小偷逃走（小偷会知道警察的布局）
思路：明显的最小割点的题目，解决这类割点问题一般都是进行拆点建图，把每个点拆成两个点（i，i+n）和一条容量为代价的边，然后道路的起点就是该起点的出点（x+n），终点就是该终点的入点（y），容量为inf无限大，这样就不会去割边了，最后以为最小割等于最大流，所以跑一边最大流就好了。

#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <stack>#include <cmath>#include <cstdio>#include <algorithm>#include <iomanip>#define N 40010#define M 100010 #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 = 1000000007;int head[N], level[N];int n, m, cnt;struct node {    int to;    int cap;//剩余流量    int next;}edge[2 * M];struct Dinic {    void init() {        memset(head, -1, sizeof(head));        cnt = 0;    }    void add(int u, int v, int cap) {//有向图        edge[cnt].to = v, edge[cnt].cap = cap, edge[cnt].next = head[u], head[u] = cnt++;        edge[cnt].to = u, edge[cnt].cap = 0, edge[cnt].next = head[v], head[v] = cnt++;//反向边    }    bool bfs(int s, int t) {//建立分层图        memset(level, -1, sizeof(level));        queue<int>q;        level[s] = 0;//源点的层次最高        q.push(s);        while (!q.empty()) {            int u = q.front();            q.pop();            for (int i = head[u]; i != -1; i = edge[i].next) {                int v = edge[i].to;                if (edge[i].cap > 0 && level[v] < 0) {                    level[v] = level[u] + 1;                    q.push(v);                }            }        }        return level[t] != -1;    }    int dfs(int u, int t, int num) {//找增广路        if (u == t) {//找到了汇点返回当前的最小值，在这条路径上分别减去最小值            return num;        }        for (int i = head[u]; i != -1; i = edge[i].next) {            int v = edge[i].to;            if (edge[i].cap > 0 && level[u] < level[v]) {                int d = dfs(v, t, min(num, edge[i].cap));                if (d > 0) {                    edge[i].cap -= d;                    edge[i ^ 1].cap += d;//反向边加值                    return d;                }            }        }        level[u] = -1;        return 0;    }    int minflow(int s, int t) {//源点和汇点        int sum = 0, num;        while (bfs(s, t)) {            while (num = dfs(s, t, inf), num > 0) {//当前层次图不断的找增广路                sum += num;            }        }        return sum;    }}dc;int main() {    cin.sync_with_stdio(false);    int T;    int a, b;    int s, t;    cin >> T;    while (T--) {        dc.init();        cin >> n >> m >> s >> t;        dc.add(s, s + n, inf);        dc.add(t, t + n, inf);        for (int i = 1; i <= n; i++) {            cin >> a;            dc.add(i, i + n, a);        }        for (int i = 0; i < m; i++) {            cin >> a >> b;            dc.add(a + n, b, inf);            dc.add(b + n, a, inf);        }        cout << dc.minflow(s, t + n) << endl;    }    return 0;}