Control

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2247    Accepted Submission(s): 940

Problem Description

　
　You, the head of Department of Security, recently received a top-secret
information that a group of terrorists is planning to transport some
WMD ¹ from one city (the source) to another one (the
destination). You know their date, source and destination, and they are
using the highway network.
　　The highway network consists of
bidirectional highways, connecting two distinct city. A vehicle can only
enter/exit the highway network at cities only.
　　You may locate some
SA (special agents) in some selected cities, so that when the
terrorists enter a city under observation (that is, SA is in this city),
they would be caught immediately.
　　It is possible to locate SA in
all cities, but since controlling a city with SA may cost your
department a certain amount of money, which might vary from city to
city, and your budget might not be able to bear the full cost of
controlling all cities, you must identify a set of cities, that:
　　* all traffic of the terrorists must pass at least one city of the set.
　　* sum of cost of controlling all cities in the set is minimal.
　　You may assume that it is always possible to get from source of the terrorists to their destination.
------------------------------------------------------------
¹ Weapon of Mass Destruction

Input

　　There are several test cases.
　
　The first line of a single test case contains two integer N and M ( 2
<= N <= 200; 1 <= M <= 20000), the number of cities and the
number of highways. Cities are numbered from 1 to N.
　　The second line contains two integer S,D ( 1 <= S,D <= N), the number of the source and the number of the destination.
　
　The following N lines contains costs. Of these lines the ith one
contains exactly one integer, the cost of locating SA in the ith city to
put it under observation. You may assume that the cost is positive and
not exceeding 10⁷.
　　The followingM lines tells you about
highway network. Each of these lines contains two integers A and B,
indicating a bidirectional highway between A and B.
　　Please process until EOF (End Of File).

Output

　　For each test case you should output exactly one line, containing one integer, the sum of cost of your selected set.
　　See samples for detailed information.

Sample Input

5 6
5 3
5
2
3
4
12
1 5
5 4
2 3
2 4
4 3
2 1

Sample Output

3

Source

2012 ACM/ICPC Asia Regional Chengdu Online

大致题意：
    给出一个又n个点，m条边组成的无向图。给出两个点s，t。对于图中的每个点，去掉这个点都需要一定的花费。求至少多少花费才能使得s和t之间不连通。

大致思路：
    最基础的拆点最大流，把每个点拆作两个点 i 和 i‘ 连接i->i‘费用为去掉这个点的花费，如果原图中有一条边a->b则连接a‘->b。对这个图求出最大流即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

const int VM=420;
const int EM=500010;
const int INF=0x3f3f3f3f;

struct Edge{
    int to,nxt;
    int cap;
}edge[EM];

int n,m,src,des,cnt,head[VM];
int dep[VM];

void addedge(int cu,int cv,int cw){
    edge[cnt].to=cv;    edge[cnt].cap=cw;   edge[cnt].nxt=head[cu];
    head[cu]=cnt++;
    edge[cnt].to=cu;    edge[cnt].cap=0;    edge[cnt].nxt=head[cv];
    head[cv]=cnt++;
}

int BFS(){
    queue<int> q;
    while(!q.empty())
        q.pop();
    memset(dep,-1,sizeof(dep));
    dep[src]=0;
    q.push(src);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=head[u];i!=-1;i=edge[i].nxt){
            int v=edge[i].to;
            if(edge[i].cap>0 && dep[v]==-1){
                dep[v]=dep[u]+1;
                q.push(v);
            }
        }
    }
    return dep[des]!=-1;
}

int DFS(int u,int minx){
    if(u==des)
        return minx;
    int tmp;
    for(int i=head[u];i!=-1;i=edge[i].nxt){
        int v=edge[i].to;
        if(edge[i].cap>0 && dep[v]==dep[u]+1 && (tmp=DFS(v,min(minx,edge[i].cap)))){
            edge[i].cap-=tmp;
            edge[i^1].cap+=tmp;
            return tmp;
        }
    }
    dep[u]=-1;
    return 0;
}

int Dinic(){
    int ans=0,tmp;
    while(BFS()){
        while(1){
            tmp=DFS(src,INF);
            if(tmp==0)
                break;
            ans+=tmp;
        }
    }
    return ans;
}

int main(){
    int s,t;
    while(~scanf("%d%d",&n,&m)){
        cnt=0;
        memset(head,-1,sizeof(head));
        scanf("%d%d",&s,&t);
        src=0, des=2*n+1;
        addedge(src,s,INF);
        addedge(n+t,des,INF);
        int u,v,w;
        for(int i=1;i<=n;i++){
            scanf("%d",&w);
            addedge(i,n+i,w);
            addedge(n+i,i,w);
        }
        for(int i=1;i<=m;i++){
            scanf("%d%d",&u,&v);
            addedge(n+u,v,INF);     //注意这里的建边,src--->s--->u(某条边)---->n+u(拆分u点后的另一点)---->v---->n+v(拆分v点后的另一点)---->u-----
            addedge(n+v,u,INF);     //所以，addedge(n+u,v,INF);仔细想想，这样才能保证 u 和 v 使连接着的
        }
        printf("%d\n",Dinic());
    }
    return 0;
}