2014多校联合第十场A题||hdu 4971 最小割定理在最大权闭合图上的应用

http://acm.hdu.edu.cn/showproblem.php?pid=4971

Problem Description

There's a company with several projects to be done. Finish a project will get you profits. However, there are some technical problems for some specific projects. To solve the problem, the manager will train his employee which may cost his budget. There may
be dependencies between technical problems, for example, A requires B means you need to solve problem B before solving problem A. If A requires B and B requires A, it means that you should solve them at the same time. You can select which problems to be solved
and how to solve them freely before finish your projects. Can you tell me the maximum profit?

 

Input

The first line of the input is a single integer T(<=100) which is the number of test cases. 

Each test case contains a line with two integer n(<=20) and m(<=50) which is the number of project to select to complete and the number of technical problem.

Then a line with n integers. The i-th integer(<=1000) means the profit of complete the i-th project.

Then a line with m integers. The i-th integer(<=1000) means the cost of training to solve the i-th technical problem.

Then n lines. Each line contains some integers. The first integer k is the number of technical problems, followed by k integers implying the technical problems need to solve for the i-th project.

After that, there are m lines with each line contains m integers. If the i-th row of the j-th column is 1, it means that you need to solve the i-th problem before solve the j-th problem. Otherwise the i-th row of the j-th column is 0.

 

Output

For each test case, please output a line which is "Case #X: Y ", X means the number of the test case and Y means the the maximum profit.

 

Sample Input

4
2 3
10 10
6 6 6
2 0 1
2 1 2
0 1 0
1 0 0
0 0 0
2 3
10 10
8 10 6
1 0
1 2
0 1 0
1 0 0
0 0 0
2 3
10 10
8 10 6
1 0
1 2
0 1 0
0 0 0
0 0 0
2 3
10 10
8 10 6
1 0
1 2
0 0 0
1 0 0
0 0 0

 

Sample Output

Case #1: 2
Case #2: 4
Case #3: 4
Case #4: 6

    悲哀的是到今天才完成多校联合的题目==从这道题开始我再写博客的时候不在写题目的大意了，太浪费青春了。。。

解题思路：最大权闭合图问题，经典解法见胡伯涛的论文，最小割模型在信息学竞赛中的应用，有一个公式：最大权闭合图的权值=所有的正权边 - 最小割。

代码如下:

#include<cstdio>
#include<iostream>
using namespace std;
const int oo=1e9;

const int mm=111111;

const int mn=999;

int node,src,dest,edge;

int ver[mm],flow[mm],next[mm];

int head[mn],work[mn],dis[mn],q[mn];
void prepare(int _node,int _src,int _dest)
{
node=_node,src=_src,dest=_dest;
for(int i=0; i<node; ++i)head[i]=-1;
edge=0;
}
void addedge(int u,int v,int c)
{
ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++;
ver[edge]=u,flow[edge]=0,next[edge]=head[v],head[v]=edge++;
}
bool Dinic_bfs()
{
int i,u,v,l,r=0;
for(i=0; i<node; ++i)dis[i]=-1;
dis[q[r++]=src]=0;
for(l=0; l<r; ++l)
for(i=head[u=q[l]]; i>=0; i=next[i])
if(flow[i]&&dis[v=ver[i]]<0)
{
dis[q[r++]=v]=dis[u]+1;
if(v==dest)return 1;
}
return 0;
}
int Dinic_dfs(int u,int exp)
{
if(u==dest)return exp;
for(int &i=work[u],v,tmp; i>=0; i=next[i])
if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
{
flow[i]-=tmp;
flow[i^1]+=tmp;
return tmp;
}
return 0;
}
int Dinic_flow()
{
int i,ret=0,delta;
while(Dinic_bfs())
{
for(i=0; i<node; ++i)work[i]=head[i];
while(delta=Dinic_dfs(src,oo))ret+=delta;
}
return ret;
}
int n,m;
int main()
{
int T,tt=0;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
prepare(n+m+2,0,n+m+1);
int x,sum=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&x);
sum+=x;
addedge(src,i,x);
}
for(int i=n+1;i<=n+m;i++)
{
scanf("%d",&x);
addedge(i,dest,x);
}
int k;
for(int i=1;i<=n;i++)
{
scanf("%d",&k);
while(k--)
{
scanf("%d",&x);
addedge(i,x+n+1,oo);
}
}
for(int i=n+1;i<=m+n;i++)
for(int j=n+1;j<=m+n;j++)
{
scanf("%d",&x);
if(x&1)
addedge(i,j,oo);
}
printf("Case #%d: %d\n",++tt,sum-Dinic_flow());
}
return 0;
}