hdu 4289 Control 最大流+拆点

Control

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4727    Accepted Submission(s): 1962

 

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.
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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：

       n个城市，m条双向边。现在有一群盗贼想从城市S到城市D，每个城市你都可以安排特工（当然需要一定的花费，特工不会白干活），当盗贼经过这个城市的时候就会被抓住，现在需要你花最少的钱安排特工，使得盗贼不管怎么走都会被抓住。

 

做法：

       普通的最大流是边上带权，但是这道题的权值在点上，所以我们需要拆点把点拆开，当作是经过这个点就要花费这么多权值。本题把花费当作流量，就是要求最小割，因为最小割等于最大流，所以就是求最大流。  

        拆点就是把该点变成，边的权值就是该点的花费，表示如果要走到这个点，则一定要有这样的花费，再把a到b的边变成 和，表示如果要从b到a的话必须要先经过b点（即已经到b'）。然后就是丢进dinic跑最大流。

---

#include<vector>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int Ni = 200005;
const int MAX = 1<<26;
struct Edge{
    int u,v,c;
    int next;
}edge[3*Ni];
int n,m;
int edn;//边数
int p[Ni];//父亲
int d[Ni];
int sp,tp;//原点，汇点

void addedge(int u,int v,int c){
    edge[edn].u=u; edge[edn].v=v; edge[edn].c=c;
    edge[edn].next=p[u]; p[u]=edn++;

    edge[edn].u=v; edge[edn].v=u; edge[edn].c=0;
    edge[edn].next=p[v]; p[v]=edn++;
}
int bfs(){
    queue <int> q;
    memset(d,-1,sizeof(d));
    d[sp]=0;
    q.push(sp);
    while(!q.empty()){
        int cur=q.front();
        q.pop();
        for(int i=p[cur];i!=-1;i=edge[i].next){
            int u=edge[i].v;
            if(d[u]==-1 && edge[i].c>0){
                d[u]=d[cur]+1;
                q.push(u);
            }
        }
    }
    return d[tp] != -1;
}
int dfs(int a,int b){
    int r=0;
    if(a==tp)return b;
    for(int i=p[a];i!=-1 && r<b;i=edge[i].next)
    {
        int u=edge[i].v;
        if(edge[i].c>0 && d[u]==d[a]+1)
        {
            int x=min(edge[i].c,b-r);
            x=dfs(u,x);
            r+=x;
            edge[i].c-=x;
            edge[i^1].c+=x;
        }
    }
    if(!r)d[a]=-2;
    return r;
}
int gainp(int x,int y){
    return 2*x-y;
}
int dinic(int sp,int tp){
    int total=0,t;
    while(bfs()){
        while(t=dfs(sp,MAX))
        total+=t;
    }
    return total;
}
int main(){
    int i,u,v,c,x;
    while(~scanf("%d%d",&n,&m))
    {
        edn=0;//初始化
        memset(p,-1,sizeof(p));
        sp=0,tp=n*2+2;
        scanf("%d%d",&u,&v);
        addedge(sp,gainp(u,0),MAX);
        addedge(gainp(v,1),tp,MAX);
        for(int i=1;i<=n;i++){
            scanf("%d",&c);
            addedge(gainp(i,0),gainp(i,1),c);
        }
        for(int i=0;i<m;i++){
            scanf("%d%d",&u,&v);
            addedge(gainp(u,1),gainp(v,0),MAX);
            addedge(gainp(v,1),gainp(u,0),MAX);
        }
        printf("%d\n",dinic(sp,tp));
    }
    return 0;
}