http://poj.org/problem?id=2195
题意:m表示人,H表示房子,一个人只能进一个房子,一个房子也只能进去一个人,房子数等于人数,现在要让所有人进入房子,求所有人都进房子最短的路径。
题解:费用流。取超级源点s=0,超级汇点t=N;s与人建边,房子与t建边,这些边费用为0容量为1;然后人与房子建边,容量为1费用为距离。
代码:
#include <iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=400;
const int maxm=50005;
const int inf=0x3f3f3f3f;
struct Edge{
int to,next,cap,flow,cost;
}edge[maxm];
int tol;
int head[maxn];
int pre[maxn],dis[maxn];
bool vis[maxn];
int N;
void init(int n){
N=n;
tol=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost){
edge[tol].to=v;
edge[tol].cost=cost;
edge[tol].cap=cap;
edge[tol].flow=0;
edge[tol].next=head[u];
head[u]=tol++;
edge[tol].to=u;
edge[tol].cost=-cost;
edge[tol].cap=0;
edge[tol].flow=0;
edge[tol].next=head[v];
head[v]=tol++;
}
bool spfa(int s,int t){
queue<int>q;
int i;
for(i=0;i<N;i++){
dis[i]=inf;
vis[i]=false;
pre[i]=-1;
}
dis[s]=0;
vis[s]=true;
q.push(s);
while(!q.empty()){
int u=q.front();
q.pop();
vis[u]=false;
for(i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].to;
if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost){
dis[v]=dis[u]+edge[i].cost;
pre[v]=i;
if(!vis[v]){
vis[v]=true;
q.push(v);
}
}
}
}
if(pre[t]==-1)return false;
else return true;
}
int mincostmaxflow(int s,int t,int &cost){
int flow=0;
cost=0;
while(spfa(s,t)){
int Min=inf;
int i;
for(i=pre[t];i!=-1;i=pre[edge[i^1].to]){
if(Min>edge[i].cap-edge[i].flow)Min=edge[i].cap-edge[i].flow;
}
for(i=pre[t];i!=-1;i=pre[edge[i^1].to]){
edge[i].flow+=Min;
edge[i^1].flow-=Min;
cost+=edge[i].cost*Min;
}
flow+=Min;
}
return flow;
}
char mp[105][105];
struct node{
int x, y;
}man[105],house[105];
int myabs(int x){
return x>0?x:-x;
}
int len(node a,node b){
return myabs(a.x-b.x)+myabs(a.y-b.y);
}
int main()
{
int n,m,i,j;
int num_man,num_house;
while(~scanf("%d%d",&n,&m)){
if(n==0&&m==0)break;
num_man=0;
num_house=0;
for(i=0;i<n;i++){
for(j=0;j<m;j++){
cin>>mp[i][j];
if(mp[i][j]=='m'){
man[num_man].x=i;
man[num_man].y=j;
num_man++;
}
else if(mp[i][j]=='H'){
house[num_house].x=i;
house[num_house].y=j;
num_house++;
}
}
}
init(2*num_house+2);
for(i=0;i<num_man;i++){
addedge(0,i+1,1,0);
for(j=0;j<num_house;j++){
addedge(i+1,j+num_house+1,1,len(man[i],house[j]));
}
}
for(i=0;i<num_house;i++)addedge(i+num_house+1,2*num_house+1,1,0);
int mincost;
mincostmaxflow(0,2*num_house+1,mincost);
printf("%d\n",mincost);
}
return 0;
}