kuangbin专题专题十一 网络流 Going Home POJ - 2195

题目链接：https://vjudge.net/problem/POJ-2195

思路：曼哈顿距离来求每个人到每个房间的距离，把距离当作费用。

就可以用最小费用最大流来解决了，把每个房子拆成两个点，限流。

源点->人->房入->房出->汇点。流量的话都设置为1，起到限流作用。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <cmath>
 using namespace std;

 const int N = ,INF = (int)1e9;
 int n,m,tot,num;
 int head[N<<],d[N<<],vis[N<<],pre[N<<];
 char mp[N][N];
 vector<pair<int,int > > p;
 vector<pair<int,int > > h;
 struct node{
     int to,nxt,cap,flow,cost;
 }e[N*N];

 void show(){
     cout << endl;
     for(int i = ; i < n; ++i) cout << mp[i] << endl;
     for(int i = ; i < num; ++i) printf("( %d, %d) ",p[i].first,p[i].second);
     cout << endl;
     for(int i = ; i < num; ++i) printf("( %d, %d) ",h[i].first,h[i].second);
     cout << endl << endl;
 }
 //求距离
 inline int _dis(int x,int y){
     return abs(p[x].first - h[y].first) + abs(p[x].second - h[y].second);
 }

 inline void add(int u,int v,int cap,int flow,int cost){
     e[tot].to = v;
     e[tot].cap = cap;
     e[tot].flow = flow;
     e[tot].cost = cost;
     e[tot].nxt = head[u];
     head[u] = tot++;
     e[tot].to = u;
     e[tot].cap = ;
     e[tot].flow = flow;
     e[tot].cost = -cost;
     e[tot].nxt = head[v];
     head[v] = tot++;
 }

 void build_map(int s,int t){

     //0源点 1~num人  num+1~2*num 房入 2*num+1~3*num房出 3*num+1汇点
     int cost;
     for(int i = ; i < num; ++i){
         for(int j = ; j < num; ++j){
             cost = _dis(i,j);
             add(i+,j++num,,,cost);
         }
     }
     for(int i = ; i < num; ++i) add(s,i+,,,);
     for(int i = ; i < num; ++i) add(i++num,i++*num,,,);
     for(int i = ; i < num; ++i) add(i++*num,t,,,);
 }

 bool spfa(int s,int t){
     for(int i = s; i <= t; ++i) pre[i] = -;
     for(int i = s; i <= t; ++i) d[i] = INF; d[s] = ;
     for(int i = s; i <= t; ++i) vis[i] = false; vis[s] = true;
     queue<int > que;
     que.push(s);
     while(!que.empty()){
         int now = que.front(); que.pop();
         vis[now] = false;
         for(int o = head[now]; ~o; o = e[o].nxt){
             int to = e[o].to;
             if(e[o].cap > e[o].flow && d[to] > d[now] + e[o].cost){
                 d[to] = d[now] + e[o].cost;
                 pre[to] = o;
                 if(!vis[to])
                     vis[to] = true;
                     que.push(to);
             }
         }
     }
     if(pre[t] == -) return false;
     else return true;
 }

 int work(){

     int s = ,t = *num+,ans = ;
     for(int i = s; i <= t; ++i) head[i] = -; tot = ;
     build_map(s,t);
     while(spfa(s,t)){
         int Min = INF;
         for(int o = pre[t]; ~o; o = pre[e[o^].to]){
             Min = min(Min,e[o].cap - e[o].flow);
         }
         for(int o = pre[t]; ~o; o = pre[e[o^].to]){
             e[o].flow += Min;
             e[o^].flow -= Min;
         }
         ans += Min*d[t];
     }
     return ans;
 }

 int main(){

     while(~scanf("%d%d",&n,&m) && (n+m)){
         for(int i = ; i < n; ++i) scanf("%s",mp[i]);
         p.clear(); h.clear();
         for(int i = ; i < n; ++i){
             for(int j = ; j < m; ++j){
                 if(mp[i][j] == 'm') p.push_back(make_pair(i,j));
                 if(mp[i][j] == 'H') h.push_back(make_pair(i,j));
             }
         }
         num = p.size();
         //show();
         //cout <<  "------------------------------------" << work() << endl;
         cout << work() << endl;
     }

     return ;
 }