hdu 3987 Base Station 最大获利+最大权闭合图

Problem Description

A famous mobile communication company is planning to build a new set of base stations. According to the previous investigation, n places are chosen as the possible new locations to build those new stations. However, the condition of each position varies much, so the costs to built a station at different places are different. The cost to build a new station at the ith place is P _i (1<=i<=n).

When complete building, two places which both have stations can communicate with each other.

Besides, according to the marketing department, the company has received m requirements. The ith requirement is represented by three integers A _i, B _i and C _i, which means if place A _i and B _i can communicate with each other, the company will get C _i profit.

Now, the company wants to maximize the profits, so maybe just part of the possible locations will be chosen to build new stations. The boss wants to know the maximum profits.

 

 

Input

Multiple test cases (no more than 20), for each test case:
The first line has two integers n (0<n<=5000) and m (0<m<=50000).
The second line has n integers, P1 through Pn, describes the cost of each location.
Next m line, each line contains three integers, A _i, B _i and C _i, describes the ith requirement.

 

 

Output

One integer each case, the maximum profit of the company.

 

 

Sample Input

  
  

   
   5 5
1 2 3 4 5
1 2 3
2 3 4
1 3 3
1 4 2
4 5 3
  
  

 

 

Sample Output

  
  

   
   4
  
  

 

 //

 

新建一个源点S，S与每个通讯线路的点相连，边权为收益

新建一个汇点T，T与每个通讯站相连，边权为建造费用

通讯线路->通讯站之间的边权为无穷大

最后的结果为所有通讯线路总的收益-最小割

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=55010;
const int M=950000;
const int inf=(1<<28);
int head[N];
struct Edge
{
    int v,next,w;
} edge[M];
int cnt,n,s,t;//n从0开始  0->n-1
void addedge(int u,int v,int w)
{
    edge[cnt].v=v;
    edge[cnt].w=w;
    edge[cnt].next=head[u];
    head[u]=cnt++;
    edge[cnt].v=u;
    edge[cnt].w=0;
    edge[cnt].next=head[v];
    head[v]=cnt++;
}
int sap()
{
    int pre[N],cur[N],dis[N],gap[N];
    int flow=0,aug=inf,u;
    bool flag;
    for(int i=0; i<n; i++)
    {
        cur[i]=head[i];
        gap[i]=dis[i]=0;
    }
    gap[s]=n;
    u=pre[s]=s;
    while(dis[s]<n)
    {
        flag=0;
        for(int &j=cur[u]; j!=-1; j=edge[j].next)
        {
            int v=edge[j].v;
            if(edge[j].w>0&&dis[u]==dis[v]+1)
            {
                flag=1;
                if(edge[j].w<aug) aug=edge[j].w;
                pre[v]=u;
                u=v;
                if(u==t)
                {
                    flow+=aug;
                    while(u!=s)
                    {
                        u=pre[u];
                        edge[cur[u]].w-=aug;
                        edge[cur[u]^1].w+=aug;
                    }
                    aug=inf;
                }
                break;
            }
        }
        if(flag) continue;
        int mindis=n;
        for(int j=head[u]; j!=-1; j=edge[j].next)
        {
            int v=edge[j].v;
            if(edge[j].w>0&&dis[v]<mindis)
            {
                mindis=dis[v];
                cur[u]=j;
            }
        }
        if((--gap[dis[u]])==0)
            break;
        gap[dis[u]=mindis+1]++;
        u=pre[u];
    }
    return flow;
}

//初始化  cnt=0;memset(head,-1,sizeof(head));

int main()
{
    int m,p;
    while(scanf("%d%d",&m,&p)==2)
    {
        cnt=0;
        memset(head,-1,sizeof(head));
        n=m+p+2;
        s=0,t=n-1;
        int sum=0;
        for(int i=1;i<=m;i++)
        {
            int cost;scanf("%d",&cost);
            addedge(i,t,cost);
        }
        for(int i=1;i<=p;i++)
        {
            int u,v,w;scanf("%d%d%d",&u,&v,&w);
            int h=m+i;
            sum+=w;
            addedge(s,h,w);
            addedge(h,u,inf);
            addedge(h,v,inf);
        }
        printf("%d\n",sum-sap());
    }
    return 0;
}