HDU - 4289 Control 最大流， 拆点， 最小割

题意：有一个恐怖分子要从a点到b点，他可能通过所有线路。在一些点中放置特工，使一定抓到恐怖分子。特工在每个点的价格不一样，求最小总价格。

把每一个点拆点，i->i'流量为权值，连接的点按拆点后连接即可。

gg啊，无向图没看到wa了好几发。。。

链接：hdu- 4289

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define INF 0x3f3f3f3f
#define ll long long

using namespace std;

const int inf = 0x3f3f3f3f;
const int maxn = 200005;

int n, m;//点数、边数
int sp, tp;//原点、汇点

struct node {
    int v, next;
    ll cap;
}mp[maxn];

int pre[maxn], cur[maxn];//cur为当前弧优化，dis存储分层图中每个点的层数（即到原点的最短距离），pre建邻接表
ll dis[maxn];
int cnt = 0;

void init() {  //不要忘记初始化
    cnt = 0;
    memset(pre, -1, sizeof(pre));
}

void add(int u, int v, ll w) { //加边
    mp[cnt].v = v;
    mp[cnt].cap = w;
    mp[cnt].next = pre[u];
    pre[u] = cnt++;
    mp[cnt].v = u;
    mp[cnt].cap = 0;
    mp[cnt].next = pre[v];
    pre[v] = cnt++;
}

bool bfs() {  //建分层图
    memset(dis, -1, sizeof(dis));
    queue<int>q;
    while(!q.empty())
        q.pop();
    q.push(sp);
    dis[sp] = 0;
    int u, v;
    while(!q.empty()) {
        u = q.front();
        q.pop();
        for(int i = pre[u]; i != -1; i = mp[i].next) {
            v = mp[i].v;
            if(dis[v] == -1 && mp[i].cap > 0) {
                dis[v] = dis[u] + 1;
                q.push(v);
                if(v == tp)
                    break;
            }
        }
    }
    return dis[tp] != -1;
}

ll dfs(int u, ll cap) {//寻找增广路
    if(u == tp || cap == 0)
    return cap;
    ll res = 0, f;
    for(int i = cur[u]; i != -1; i = mp[i].next) {//
        int v = mp[i].v;
        if(dis[v] == dis[u] + 1 && (f = dfs(v, min(cap - res, mp[i].cap))) > 0) {
            mp[i].cap -= f;
            mp[i ^ 1].cap += f;
            res += f;
            if(res == cap)
                return cap;
        }
    }
    if(!res)
        dis[u] = -1;
    return res;
}

ll dinic() {
    ll ans = 0;
    while(bfs()) {
        for(int i = 0; i <= 2 * n + 1; i++)
            cur[i] = pre[i];
        ans += dfs(sp, inf);
    }
    return ans;
}

int main ()
{
    while(~scanf("%d %d", &n, &m)) {
        init();
        scanf("%d %d", &sp, &tp);
        tp += n;
        ll p;
        for(int i = 1; i <= n; i++) {
            scanf("%lld", &p);
            add(i, i + n, p);
        }
        int s, t;
        for(int i = 1; i <= m; i++) {
            scanf("%d %d", &s, &t);
            add(s + n, t, inf);
            add(t + n, s, inf);
        }
        ll res = dinic();
        printf("%lld\n", res);
    }
    return 0;
}