题目

Source

http://acm.hdu.edu.cn/showproblem.php?pid=4307

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

分析

化一下那个矩阵，可以知道，目标是最大化这个：

$$D = \sum_{i=1}^N \sum_{j=1}^N A_iA_jB_{ij} - \sum_{i=1}^NC_i$$

这样就是100多W个点的最大权闭合子图了= =。。

其实式子还可以再化。。 http://blog.csdn.net/weiguang_123/article/details/8077385。

然后就化成了最小割的经典二者选一的模型了，POJ3469。

这我一点都化不来= =。。

代码

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define MAXN 1111
#define MAXM 2222*1111

struct Edge{
    int v,cap,flow,next;
}edge[MAXM];
int vs,vt,NE,NV;
int head[MAXN];

void addEdge(int u,int v,int cap){
    edge[NE].v=v; edge[NE].cap=cap; edge[NE].flow=0;
    edge[NE].next=head[u]; head[u]=NE++;
    edge[NE].v=u; edge[NE].cap=0; edge[NE].flow=0;
    edge[NE].next=head[v]; head[v]=NE++;
}

int level[MAXN];
int gap[MAXN];
void bfs(){
    memset(level,-1,sizeof(level));
    memset(gap,0,sizeof(gap));
    level[vt]=0;
    gap[level[vt]]++;
    queue<int> que;
    que.push(vt);
    while(!que.empty()){
        int u=que.front(); que.pop();
        for(int i=head[u]; i!=-1; i=edge[i].next){
            int v=edge[i].v;
            if(level[v]!=-1) continue;
            level[v]=level[u]+1;
            gap[level[v]]++;
            que.push(v);
        }
    }
}

int pre[MAXN];
int cur[MAXN];
int ISAP(){
    bfs();
    memset(pre,-1,sizeof(pre));
    memcpy(cur,head,sizeof(head));
    int u=pre[vs]=vs,flow=0,aug=INF;
    gap[0]=NV;
    while(level[vs]<NV){
        bool flag=false;
        for(int &i=cur[u]; i!=-1; i=edge[i].next){
            int v=edge[i].v;
            if(edge[i].cap!=edge[i].flow && level[u]==level[v]+1){
                flag=true;
                pre[v]=u;
                u=v;
                //aug=(aug==-1?edge[i].cap:min(aug,edge[i].cap));
                aug=min(aug,edge[i].cap-edge[i].flow);
                if(v==vt){
                    flow+=aug;
                    for(u=pre[v]; v!=vs; v=u,u=pre[u]){
                        edge[cur[u]].flow+=aug;
                        edge[cur[u]^1].flow-=aug;
                    }
                    //aug=-1;
                    aug=INF;
                }
                break;
            }
        }
        if(flag) continue;
        int minlevel=NV;
        for(int i=head[u]; i!=-1; i=edge[i].next){
            int v=edge[i].v;
            if(edge[i].cap!=edge[i].flow && level[v]<minlevel){
                minlevel=level[v];
                cur[u]=i;
            }
        }
        if(--gap[level[u]]==0) break;
        level[u]=minlevel+1;
        gap[level[u]]++;
        u=pre[u];
    }
    return flow;
}

int B[1111][1111],C[1111];

int main(){
	int t,n;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		int sum=0;
		for(int i=1; i<=n; ++i){
			for(int j=1; j<=n; ++j){
				scanf("%d",&B[i][j]);
				sum+=B[i][j];
			}
		}
		for(int i=1; i<=n; ++i){
			scanf("%d",&C[i]);
		}
		vs=0; vt=n+1; NV=vt+1; NE=0;
		memset(head,-1,sizeof(head));
		for(int i=1; i<=n; ++i){
			addEdge(vs,i,C[i]);
			int tmp=0;
			for(int j=1; j<=n; ++j) tmp+=B[i][j];
			addEdge(i,vt,tmp);
			for(int j=1; j<=n; ++j){
				if(i!=j) addEdge(j,i,B[i][j]);
			}
		}
		printf("%d\n",sum-ISAP());
	}
	return 0;
}