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poj2195 Going Home(最大费用最小流)

2016-07-22 20:12 459 查看
题目链接:

http://poj.org/problem?id=2195

求最大流最小费用。主要是加边,

先从源点到各个人m,费用为0流量为1,。

然后从m到H,费用为距离,流量为1.

最后从H到汇点,费用为0,流量为1.

代码:

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<stdlib.h>
#define inf 0x3f3f3f3f
#define MM 10005
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
struct Edge{
int from,to,cap,flow,cost,next;
Edge(){}
Edge(int from,int to,int cap,int flow,int cost,int next){
this->from=from;
this->to=to;
this->cap=cap;
this->flow=flow;
this->cost=cost;
this->next=next;
}
}ed[666666];
char str[105][105];
int n,m,s,t,head,tail,lnum;
int vis[MM],dis[MM],pre[MM];//pre 上一条弧
int cag[MM],que[MM*3],start[MM];//cag 可改进量
pair<int,int>house[MM],people[MM];

void init(){
lnum=0;
rep(i,s,t) start[i]=-1;
}

void add(int x,int y,int v,int c){
ed[lnum]=Edge(x,y,v,0,c,start[x]);
start[x]=lnum++;
ed[lnum]=Edge(y,x,0,0,-c,start[y]);
start[y]=lnum++;
}

bool spfa(int& flow,int& cost){
rep(i,s,t) dis[i]=inf,vis[i]=0;
head=tail=0;
dis[s]=0;vis[s]=1;pre[s]=0;cag[s]=inf;
que[tail++]=s;
while(head<tail){
int x=que[head++];
vis[x]=0;
for(int i=start[x];~i;i=ed[i].next){
Edge& e=ed[i];
if(e.cap>e.flow&&dis[e.to]>dis[x]+e.cost){
dis[e.to]=dis[x]+e.cost;
pre[e.to]=i;
cag[e.to]=min(cag[x],e.cap-e.flow);
if(!vis[e.to]){
que[tail++]=e.to;
vis[e.to]=1;
}
}
}
}
if(dis[t]==inf) return false;
flow+=cag[t];
cost+=dis[t]*cag[t];
int x=t;
while(x!=s){
ed[pre[x]].flow+=cag[t];
ed[pre[x]^1].flow-=cag[t];
x=ed[pre[x]].from;
}
return true;
}

int Mincost(){
int flow=0,cost=0;
while((spfa(flow,cost)));
return cost;
}

int main()
{
int i,j,k,tt1,tt2;
while(~scanf("%d%d",&n,&m))
{
if(n==0&&m==0)break;
init();
tt1=tt2=0;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++){
cin>>str[i][j];
if(str[i][j]=='m'){
people[++tt1]=make_pair(i,j);
}
if(str[i][j]=='H'){
house[++tt2]=make_pair(i,j);
}
}
s=0;t=tt1+tt2+1;
for(i=1;i<=tt1;i++){
add(0,i,1,0);
}
for(i=1;i<=tt1;i++)
for(j=1;j<=tt2;j++){
int d=abs(people[i].first-house[j].first)+abs(people[i].second-house[j].second);
add(i,tt1+j,1,d);
}

for(i=1;i<=tt2;i++){
add(tt1+i,t,1,0);
}

int ans =	Mincost();
printf("%d\n",ans);
}
}
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