HDU 4307 Matrix (最小割)

Matrix

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 823    Accepted Submission(s): 210

Problem Description

Let A be a 1*N matrix, and each element of A is either 0 or 1. You are to find such A that maximize D=(A*B-C)*AT, where B is a given N*N matrix whose elements are non-negative, C is a given 1*N matrix whose elements are also non-negative, and AT is the transposition of A (i.e. a N*1 matrix).

 

Input

The first line contains the number of test cases T, followed by T test cases.
For each case, the first line contains an integer N (1<=N<=1000).
The next N lines, each of which contains N integers, illustrating the matrix B. The jth integer on the ith line is B[i][j].
Then one line followed, containing N integers, describing the matrix C, the ith one for C[i].
You may assume that sum{B[i][j]} < 2^31, and sum{C[i]} < 2^31.

 

Output

For each case, output the the maximum D you may get.

 

Sample Input

  
  

   
   1
3
1 2 1
3 1 0
1 2 3
2 3 7
  
  

 

Sample Output

  
  

   
   2

   
   

    
    

     
     Hint
    
    
For sample, A=[1, 1, 0] or A=[1, 1, 1] would get the maximum D.
   
   
   
    
  
  

 

Author

BUPT

 

Source

2012 Multi-University Training Contest 1

 

Recommend

zhuyuanchen520

网上看的题解，一看是矩阵就不是很想碰的题。看过题解才知道，最小割的应用还是很强的。

参考题解出处： http://blog.youkuaiyun.com/weiguang_123/article/details/8077385

思路：

如果Ai选择0会产生sum{Bij}(1<=j<=N)的代价，如果Ai选择1会产生Ci的代价，如果Ai选择1且aj选择0就会产生Bij的代价。这样就可以建模了。

设源点和汇点，如果Ai选0则从源点到i连接一条容量为sum(Bij)的边，如果Ai选1则从i到汇点连接一条容量为Ci的边，对于从i到j的边，即Ai选了1，并且Aj选了0的边，其容量就为Bij。则从源点到汇点的最大流即最小割，那么D最大即sum(Bij)(1<=i<=N,1<=j<=N)-最小割。

#include<cstdio>
using namespace std;
const int mm=2005005;
const int mn=1005;
const int oo=1000000000;
int node,s,t,edge;
int to[mm],flow[mm],next[mm];
int head[mn],work[mn],dis[mn],q[mn];
inline int min(int a,int b)
{
    return a<b?a:b;
}
inline void init(int nn,int ss,int tt)
{
    node=nn,s=ss,t=tt,edge=0;
    for(int i=0; i<node; ++i) head[i]=-1;
}
inline void add(int u,int v,int c1,int c2=0)
{
    to[edge]=v,flow[edge]=c1,next[edge]=head[u],head[u]=edge++;
    to[edge]=u,flow[edge]=c2,next[edge]=head[v],head[v]=edge++;
}
bool bfs()
{
    int i,u,v,l,r=0;
    for(i=0; i<node; ++i) dis[i]=-1;
    dis[q[r++]=s]=0;
    for(l=0; l<r; ++l)
        for(i=head[u=q[l]]; i>=0; i=next[i])
            if(flow[i]&&dis[v=to[i]]<0)
            {
                dis[q[r++]=v]=dis[u]+1;
                if(v==t)return 1;
            }
    return 0;
}
int dfs(int u,int maxf)
{
    if(u==t) return maxf;
    for(int &i=work[u],v,tmp; i>=0; i=next[i])
        if(flow[i]&&dis[v=to[i]]==dis[u]+1&&(tmp=dfs(v,min(maxf,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            return tmp;
        }
    return 0;
}
long long dinic()
{
    int i,delta;
    long long ret=0;
    while(bfs())
    {
        for(i=0; i<node; ++i) work[i]=head[i];
        while(delta=dfs(s,oo))ret+=delta;
    }
    return ret;
}
int main()
{
    int i,j,n,a,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        init(n+2,0,n+1);
        long long fcnt=0;
        for(i=1; i<=n; ++i)
        {
            int ss=0;
            for(j=1; j<=n; ++j)
            {
                scanf("%d",&a),ss+=a;
                add(i,j,a);
            }
            add(s,i,ss);
            fcnt+=ss;
        }
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a);
            add(i,t,a);
        }
        printf("%I64d\n",fcnt-dinic());
    }
    return 0;
}