HDU4289:Control（最小割）

Control

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5023    Accepted Submission(s): 2067

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4289

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

题意：

给出起点和终点城市，每个城市都有一定的权值，有一群坏人要从起点到终点，现在要在一些城市上布置警察拦截这些坏人，问最少花费。

 

题解：

由于点有权值，我们考虑拆点；同时这题是双向边。

如果我们像以往那样反向边容量也为c，则有误。

我们这样考虑插边：插入的反向边容量还是0，只是这样来构造双向边：u'->v , v'->u。其中u'为出度点，u为入度点。

最后跑个最大流就行了。最大流等于最小割

 

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
#define INF 99999999
using namespace std;
typedef long long ll;
const int N = 505,M = 1e5;
int head[N],d[N];
int tot,n,m,s,t;
struct Edge{
    int v,next,c;
}e[M];
void adde(int u,int v,int c){
    e[tot].v=v;e[tot].next=head[u];e[tot].c=c;head[u]=tot++;
    e[tot].v=u;e[tot].next=head[v];e[tot].c=0;head[v]=tot++;
}
bool bfs(int S,int T){
    memset(d,0,sizeof(d));d[S]=1;
    queue <int > q;q.push(S);
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=head[u];i!=-1;i=e[i].next){
            int v=e[i].v;
            if(!d[v] && e[i].c>0){
                d[v]=d[u]+1;
                q.push(v);
            }
        }
    }
    return d[t]!=0;
}
int dfs(int u,int a){
    int flow=0,f;
    if(u==t || a==0) return a;
    for(int i=head[u];i!=-1;i=e[i].next){
        int v=e[i].v;
        if(d[v]!=d[u]+1) continue ;
        f=dfs(v,min(a,e[i].c));
        if(f>0){
            e[i].c-=f;
            e[i^1].c+=f;
            flow+=f;
            a-=f;
            if(a==0) break;
        }
    }
    if(!flow) d[u]=-1;
    return flow;
}
int Dinic(){
    int max_flow=0;
    while(bfs(s,t))
        max_flow+=dfs(s,INF);
    return max_flow;
}
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        memset(head,-1,sizeof(head));tot=0;
        scanf("%d%d",&s,&t);
        t+=200;
        for(int i=1,c;i<=n;i++){
            scanf("%d",&c);
            adde(i,i+200,c);
        }
        for(int i=1,u,v;i<=m;i++){
            scanf("%d%d",&u,&v);
            adde(u+200,v,INF);
            adde(v+200,u,INF);
        }
        printf("%d\n",Dinic());
    }
    return 0;
}