hdu-4411-Arrest

//这题不会做，看了大牛的解题报告才知道用费用流可解，不怎么熟悉，今天调试了一上午，终于弄明白了，其实最小费用流本身很好理解，最小费用在求最大流的时候和一般的网络流的盲目增广不同，而是选最小费用来增广，最后到不能增广为止，这题的难点就是怎么去构建成最小费用流；

详细思路具体参见下面这两个人的博客：

http://blog.csdn.net/flying_stones_sure/article/details/8010155

http://blog.csdn.net/kksleric/article/details/8009882

//贴个自己的代码留念一下：

#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;#define inf 1000000int map[110][110],n,m;int dis[210],vis[210];int pre[210],cap[210][210],cost[210][210];queue <int > que;int bfs(int start,int end){int i;for(i=0;i<=n;i++)dis[i]=inf;memset(vis,0,sizeof(vis));dis[start]=0;que.push(start);vis[start]=1;while(!que.empty()){int u=que.front();vis[u]=0;que.pop();for(i=0;i<=n;i++){if(cap[u][i]>0&&dis[u]+cost[u][i]<dis[i]){dis[i]=dis[u]+cost[u][i];pre[i]=u;if(!vis[i]){vis[i]=1;que.push(i);}}}}return dis[end]<inf;}int mcmf(int start,int end){int i,res=0,Min;while(bfs(start,end)){Min=inf;for(i=end;i!=start;i=pre[i])if(cap[pre[i]][i]<Min)Min=cap[pre[i]][i];for(i=end;i!=start;i=pre[i]){res+=Min*cost[pre[i]][i];cap[pre[i]][i]-=Min;cap[i][pre[i]]+=Min;}}return res;}int main(){    int a,b,c,i,j,k,g,p;    while(scanf("%d%d%d",&n,&m,&k),n+m+k)    {        for(i=0;i<=n;i++)        {            map[i][i]=0;            for(j=i+1;j<=n;j++)            {                map[i][j]=inf;                map[j][i]=inf;            }        }        for(i=1;i<=m;i++)        {            scanf("%d%d%d",&a,&b,&c);if(map[a][b]>c){map[a][b]=c;map[b][a]=c;}        }        for(g=0;g<=n;g++)        {for(i=0;i<=n;i++)            {for(j=0;j<=n;j++){if(map[i][g]+map[g][j]<map[i][j])map[i][j]=map[i][g]+map[g][j];}}}p=n;n=2*n+2;for(i=0;i<=n;i++){for(j=0;j<=n;j++){cost[i][j]=inf;cap[i][j]=0;}}for(i=1;i<=p;i++){for(j=i+1;j<=p;j++){if(map[i][j]<inf){cost[i+p+1][j]=map[i][j];cost[j][i+p+1]=-map[i][j];cap[i+p+1][j]=inf;}}cap[i][i+p+1]=1;cost[i][i+p+1]=-inf;cost[i+p+1][i]=inf;if(map[0][i]<inf){cap[i+p+1][n]=inf;cost[i+p+1][n]=map[i][0];cost[n][i+p+1]=-map[i][0];cap[p+1][i]=inf;cost[p+1][i]=map[0][i];cost[i][p+1]=-map[0][i];}}cap[0][p+1]=k;cost[0][p+1]=0;cap[p+1][n]=k;cost[p+1][n]=0;printf("%d\n",mcmf(0,n)+p*inf);    }    return 0;}/*5 5 20 1 11 2 12 3 13 5 10 4 1*/