HDU-4289 Control

題目鏈接：Control

題目大意：給出s和t點，求在圖中找出最少的點集，使s和t不聯通。

解題思路：拆點，把每個點變成一條邊，權值爲此點的花費，不同點之間若相連，則權值爲正無窮。最後跑一遍最大流就能得到解了。

代碼如下：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int, int> P;
const int inf  = 0x3f3f3f3f;
const int maxn = 2e4 + 15;

struct Edge {
    int to, cap, next;
};
int ecnt, st, ed, n, m;
Edge es[maxn * 10];
int cur[505], dep[505], head[505], cost[505]; 

class Dinic {
public:
    void init(){
        memset(head, -1, sizeof head);
        ecnt = 0;
    }

    void add_edge(int u, int v, int cap) {
        es[ecnt].to = v;
        es[ecnt].next = head[u];
        es[ecnt].cap = cap;
        head[u] = ecnt++;
    }

    void add_double(int u, int v, int w1, int w2 = 0) {
        add_edge(u, v, w1);
        add_edge(v, u, w2);
    }

    bool BFS() {
        queue<int> q; q.push(st);
        memset(dep, 0x3f, sizeof dep); dep[st] = 0;
        while(q.size() && dep[ed] == inf) {
            int u = q.front(); q.pop();
            for(int i = head[u]; ~i; i = es[i].next) {
                Edge& e = es[i];
                if(e.cap > 0 && dep[e.to] == inf) {
                    dep[e.to] = dep[u] + 1;
                    q.push(e.to);
                }
            }
        }
        return dep[ed] < inf;
    }

    int DFS(int u, int maxflow) {
        if(u == ed) return maxflow;
        int res = 0;
        for(int i = cur[u]; ~i; i = es[i].next) {
            Edge& e = es[i];
            if(dep[e.to] == dep[u] + 1 && e.cap > 0) {
                int flow = DFS(e.to, min(maxflow, e.cap));  
                cur[u] = i; 
                res += flow; maxflow -= flow;
                es[i].cap -= flow; 
                es[i ^ 1].cap += flow;
                if(!maxflow) return res;
            }
        }       
        dep[u] = inf;
        return res;
    }

    int MaxFlow(){
        int ans = 0;    
        while(BFS()) {   
            for(int i = st; i <= ed; i++) cur[i] = head[i];
            ans += DFS(st, inf);
        }
        return ans;
    }
};
inline int read() {
    char c = getchar();
    while(!isdigit(c)) c = getchar();
    int x = 0;
    while(isdigit(c)) {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x;
}

int main(){
#ifdef NEKO
    freopen("Nya.txt", "r", stdin);
#endif
    Dinic dic;
    while(scanf("%d%d", &n, &m) != EOF) {
        int s, t; 
        s = read(); t = read();
        st = 0; ed = 2 * n + 1;
        dic.init();
        dic.add_double(st, s, inf);
        dic.add_double(t + n, ed, inf);
        for(int i = 1; i <= n; i++) {
            cost[i] = read();
            dic.add_double(i, i + n, cost[i]);  
        }
        for(int i = 1; i <= m; i++) {
            int u, v; u = read(); v = read();
            dic.add_double(u + n, v, inf);  
            dic.add_double(v + n, u, inf);      
        }
        cout << dic.MaxFlow() << endl;
    }
    return 0;
}