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;

}