POJ 2195 Going Home（最小費用最大流）題解

題意：給你一張圖，有k個人和k個房子，每個房子只能住一個人，每個人到某一房子的花費爲曼哈頓距離，問你讓k個人怎麼走，使他們都住房子且花費最小。

思路：我們把所有人和超級源點相連，流量爲1花費爲0，所有房子和超級匯點相連，流量爲1花費爲0，然後把所有人和所有房子加邊，流量爲1，花費爲曼哈頓距離，這樣這道題目就變成了求超級源點到超級匯點的MCMF。

代碼：

#include<cstdio>
#include<vector>
#include<stack>
#include<queue>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#define ll long long
const int maxn = 10000+5;
const int maxm = 100000+5;
const int MOD = 1e7;
const int INF = 1 << 25;
using namespace std;
struct Edge{
    int to,next,cap,flow,cost;
}edge[maxm];
struct node{
    int x,y;
}house[maxn],man[maxn];
int head[maxn],tot;
int pre[maxn],dis[maxn];
bool vis[maxn];
int N,M;
void init(){
    N = maxn;
    tot = 0;
    memset(head,-1,sizeof(head));
}
void addEdge(int u,int v,int cap,int cost){
    edge[tot].to = v;
    edge[tot].cap = cap;    //容量
    edge[tot].flow = 0;
    edge[tot].cost = cost;
    edge[tot].next = head[u];
    head[u] = tot++;

    edge[tot].to = u;
    edge[tot].cap = 0;
    edge[tot].flow = 0;
    edge[tot].cost = -cost;
    edge[tot].next = head[v];
    head[v] = tot++;
}
bool spfa(int s,int t){
    queue<int> q;
    for(int 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(int 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);
                }
            }
        }
    }
    return pre[t] != -1;
}

int MCMF(int s,int t,int &cost){
    int flow = 0;
    cost = 0;
    while(spfa(s,t)){
        int MIN = INF;
        for(int 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(int 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[maxn][maxn];
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m) && n+m){
        init();
        int mcnt = 0,hcnt = 0;
        for(int i = 1;i <= n;i++){
            scanf("%s",mp[i] + 1);
            for(int j = 1;j <= m;j++){
                if(mp[i][j] == 'm'){
                    man[mcnt].x = i;
                    man[mcnt++].y = j;
                }
                else if(mp[i][j] == 'H'){
                    house[hcnt].x = i;
                    house[hcnt++].y = j;
                }
            }
        }

        //建圖
        for(int i = 1;i <= hcnt;i++){   //和超級源點連
            addEdge(0,i,1,0);
        }
        for(int i = hcnt + 1;i <= 2*hcnt;i++){  //和超級匯點連
            addEdge(i,2*hcnt + 1,1,0);
        }
        for(int i = 0;i < mcnt;i++){
            for(int j = 0;j < mcnt;j++){
                int pay = abs(house[i].x - man[j].x) + abs(house[i].y - man[j].y);
                addEdge(i + 1,j + mcnt + 1,1,pay);
            }
        }

        int cost = 0;
        MCMF(0,2*hcnt + 1,cost);
        printf("%d\n",cost);
    }
    return 0;
}