最大流最小割求割边

hdu3251

Being a Hero

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 867    Accepted Submission(s): 278
Special Judge

Problem Description

You are the hero who saved your country. As promised, the king will give you some cities of the country, and you can choose which ones to own!

But don't get too excited. The cities you take should NOT be reachable from the capital -- the king does not want to accidentally enter your area. In order to satisfy this condition, you have to destroy some roads. What's worse, you have to pay for that -- each road is associated with some positive cost. That is, your final income is the total value of the cities you take, minus the total cost of destroyed roads.

Note that each road is a unidirectional, i.e only one direction is available. Some cities are reserved for the king, so you cannot take any of them even if they're unreachable from the capital. The capital city is always the city number 1.

 

Input

The first line contains a single integer T (T <= 20), the number of test cases. Each case begins with three integers n, m, f (1 <= f < n <= 1000, 1 <= m < 100000), the number of cities, number of roads, and number of cities that you can take. Cities are numbered 1 to n. Each of the following m lines contains three integers u, v, w, denoting a road from city u to city v, with cost w. Each of the following f lines contains two integers u and w, denoting an available city u, with value w.

 

Output

For each test case, print the case number and the best final income in the first line. In the second line, print e, the number of roads you should destroy, followed by e integers, the IDs of the destroyed roads. Roads are numbered 1 to m in the same order they appear in the input. If there are more than one solution, any one will do.

 

Sample Input

24 4 21 2 21 3 33 2 42 4 12 34 44 4 21 2 21 3 33 2 12 4 12 34 4

 

Sample Output

Case 1: 31 4Case 2: 42 1 3

题意：给你f个可选城市，每个城市都有其价值w0,国王的城市在1,现在国王不想见到你（国王不想通过某种路径到达你选定的城市）——将你选的若干城市隔离出去，每条道路隔断都需要花费w1,现在问你可以达到的最大价值 
并要求输出你割断的路的编号 
题解：网络流模型题，将所有可选点连入超级汇点，边权为城市的价值，求出最小割（等于最大流），最大价值为所有可选点的价值和减去最小割 

所有的割边的求法：最后DFS一次，从源点出发，凡是可以到达的点进行标记，到不了的点自然就属于汇点的集合

搜索一次，属于不同集合两个点的那条边就是割边，注意此题要按照输入的顺序输出；

程序：

#include"stdio.h"#include"string.h"#include"iostream"#define M 1009#define inf 999999999using namespace std;struct st{    int u,v,w,next;}edge[400009];int head[M],dis[M],q[M],work[M],cnt[M],use[M],t;int min(int a,int b){    return a<b?a:b;}void init(){    t=0;    memset(head,-1,sizeof(head));}void add(int u,int v,int w){    edge[t].u=u;    edge[t].v=v;    edge[t].w=w;    edge[t].next=head[u];    head[u]=t++;    edge[t].u=v;    edge[t].v=u;    edge[t].w=0;    edge[t].next=head[v];    head[v]=t++;}int bfs(int S,int T){    int rear=0;    memset(dis,-1,sizeof(dis));    dis[S]=0;    q[rear++]=S;    for(int i=0;i<rear;i++)    {        for(int j=head[q[i]];j!=-1;j=edge[j].next)        {            int v=edge[j].v;            if(edge[j].w&&dis[v]==-1)            {                dis[v]=dis[q[i]]+1;                q[rear++]=v;                if(v==T)                    return 1;            }        }    }    return 0;}int dfs(int S,int a,int T){    if(S==T)        return a;    for(int &i=work[S];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(edge[i].w&&dis[v]==dis[S]+1)        {            int tt=dfs(v,min(a,edge[i].w),T);            if(tt)            {                edge[i].w-=tt;                edge[i^1].w+=tt;                return tt;            }        }    }    return 0;}int Dinic(int S,int T){    int ans=0;    while(bfs(S,T))    {        memcpy(work,head,sizeof(head));        while(int tt=dfs(S,inf,T))            ans+=tt;    }    return ans;}void DFS(int S){    use[S]=1;    for(int i=head[S];i!=-1;i=edge[i].next)    {        int v=edge[i].v;        if(!use[v]&&edge[i].w)            DFS(v);    }}int main(){    int T,m,k,n,a,b,c,kk=1,i;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&n,&m,&k);        init();        int sum=0;        while(m--)        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,c);        }        while(k--)        {            scanf("%d%d",&a,&c);            add(a,n+1,c);            sum+=c;        }        printf("Case %d: ",kk++);        int ans=Dinic(1,n+1);        printf("%d\n",sum-ans);        memset(use,0,sizeof(use));        DFS(1);        int r=0;        for(i=0;i<t;i+=2)        {            int u=edge[i].u;            int v=edge[i].v;            if(use[u]&&!use[v]&&v!=n+1)                cnt[r++]=i/2+1;        }        printf("%d",r);        for(i=0;i<r;i++)            printf(" %d",cnt[i]);        printf("\n");    }    return 0;}