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
",sum-ISAP());
}
return 0;
}