hdu3523(费用流)

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题意：给出一个n*m的矩阵，要求最后价值最小，价值为找出一行1~m的序列，并用每列点的权值减去序列的权值绝对值的和

代码：#include <math.h>
#include <vector>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
struct node{
int to,cap,cost,rev;
};
int V;
int dis[10005],prevv[10005],preve[10005];
vector<node> G[10005];
void addedge(int from,int to,int cap,int cost){
G[from].push_back((node){to,cap,cost,G[to].size()});
G[to].push_back((node){from,0,-cost,G[from].size()-1});
}
int min_cost_flow(int s,int t,int f){
int i,d,v,ans,sign;
ans=0;
while(f>0){
fill(dis,dis+V,INF);
dis[s]=0;
sign=1;
while(sign){
sign=0;
for(v=0;v<V;v++){
if(dis[v]==INF)
continue;
for(i=0;i<G[v].size();i++){
node &e=G[v][i];
if(e.cap>0&&dis[e.to]>dis[v]+e.cost){
dis[e.to]=dis[v]+e.cost;
prevv[e.to]=v;
preve[e.to]=i;
sign=1;
}
}
}
}
if(dis[t]==INF)
return -1;
d=f;
for(v=t;v!=s;v=prevv[v])
d=min(d,G[prevv[v]][preve[v]].cap);
f-=d;
ans+=d*dis[t];
for(v=t;v!=s;v=prevv[v]){
node &e=G[prevv[v]][preve[v]];
e.cap-=d;
G[v][e.rev].cap+=d;
}
}
return ans;
}
int g[105][105],sum[105][105];
int main(){
int i,j,k,t,n,m,ans,cas;
scanf("%d",&t);
for(cas=1;cas<=t;cas++){
scanf("%d%d",&m,&n);
V=2*m+2;
for(i=0;i<=2*m+2;i++)
G[i].clear();
memset(sum,0,sizeof(sum));
for(i=1;i<=n;i++)
for(j=1;j<=m;j++){
scanf("%d",&g[i][j]);
for(k=1;k<=m;k++)
sum[j][k]+=abs(g[i][j]-k); //sum[i][j]代表dii列减j的价值
}
for(i=1;i<=m;i++){ //左边为m列，右边为1~m
addedge(0,i,1,0);
addedge(i+m,2*m+1,1,0);
}
for(i=1;i<=m;i++)
for(j=1;j<=m;j++)
addedge(i,j+m,1,sum[i][j]);
ans=min_cost_flow(0,2*m+1,m); //求一个流量是m的费用流
printf("Case #%d: %d\n",cas,ans);
}
return 0;
}