kuangbin专题专题十一 网络流 Going Home POJ

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 = 510,INF = (int)1e9;
int n,m,tot,num;
int head[N<<2],d[N<<2],vis[N<<2],pre[N<<2];
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 = 0; i < n; ++i) cout << mp[i] << endl;
    for(int i = 0; i < num; ++i) printf("( %d, %d) ",p[i].first,p[i].second);
    cout << endl;
    for(int i = 0; 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 = 0;
    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 = 0; i < num; ++i){
        for(int j = 0; j < num; ++j){
            cost = _dis(i,j);
            add(i+1,j+1+num,1,0,cost);
        }
    }
    for(int i = 0; i < num; ++i) add(s,i+1,1,0,0);
    for(int i = 0; i < num; ++i) add(i+1+num,i+1+2*num,1,0,0);
    for(int i = 0; i < num; ++i) add(i+1+2*num,t,1,0,0);
}

bool spfa(int s,int t){
    for(int i = s; i <= t; ++i) pre[i] = -1;
    for(int i = s; i <= t; ++i) d[i] = INF; d[s] = 0;
    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] == -1) return false;
    else return true;
}

int work(){

    int s = 0,t = 3*num+1,ans = 0;
    for(int i = s; i <= t; ++i) head[i] = -1; tot = 0;
    build_map(s,t);
    while(spfa(s,t)){
        int Min = INF;
        for(int o = pre[t]; ~o; o = pre[e[o^1].to]){
            Min = min(Min,e[o].cap - e[o].flow);
        }
        for(int o = pre[t]; ~o; o = pre[e[o^1].to]){
            e[o].flow += Min;
            e[o^1].flow -= Min;
        }
        ans += Min*d[t];
    }
    return ans;
}

int main(){

    while(~scanf("%d%d",&n,&m) && (n+m)){
        for(int i = 0; i < n; ++i) scanf("%s",mp[i]);
        p.clear(); h.clear();
        for(int i = 0; i < n; ++i){
            for(int j = 0; 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 0;
}