Going Home
On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man.
Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a '.' means an empty space, an 'H' represents a house on that point, and am 'm' indicates there is a little man on that point.
You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.
Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a '.' means an empty space, an 'H' represents a house on that point, and am 'm' indicates there is a little man on that point.
You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.
InputThere are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both N and M are between 2 and 100, inclusive. There will be the same number of 'H's and 'm's on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.
OutputFor each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.
Sample Input
2 2 .m H. 5 5 HH..m ..... ..... ..... mm..H 7 8 ...H.... ...H.... ...H.... mmmHmmmm ...H.... ...H.... ...H.... 0 0
Sample Output
2 10 28
给你一张图，图上有若干个人和若干个屋子，现在要使的这若干个人都进到屋子里，并且一个屋子只能进一个人，求总步数最小。
思路：图转边 建图（超级源点和汇点）跑最小费用最大流板子即可
#include<bits/stdc++.h> #define FIN freopen("input.txt","r",stdin) #define ll long long #define mod 1000000007 #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define inf 0x3f3f3f3f const int maxn = 50005; using namespace std; int s,t,cnt=1,n,m; char str[205][205]; int head[maxn],Next[maxn],To[maxn],ct[maxn],flow[maxn],dis[maxn],vis[maxn],pre[maxn],id[maxn]; struct node{ int x; int y; }p[maxn]; void add(int u,int v,int w,int cost){ Next[++cnt]=head[u]; head[u]=cnt; To[cnt]=v; flow[cnt]=w; ct[cnt]=cost; } void get_graph(){ s=0; t=0; int hh=0,mm=0; for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ if(str[i][j]=='H'){ hh++; p[hh].x=i; p[hh].y=j; } } } for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ if(str[i][j]=='m'){ mm++; add(s,mm+hh,1,0); add(mm+hh,s,0,0); for(int k=1;k<=hh;k++){ add(mm+hh,k,1,abs(i-p[k].x)+abs(j-p[k].y)); add(k,mm+hh,0,-abs(i-p[k].x)-abs(j-p[k].y)); } } } } t=hh+mm+1; for(int i=1;i<=hh;i++){ add(i,t,1,0); add(t,i,0,0); } } bool spfa(){ memset(vis,0,sizeof(vis)); memset(dis,inf,sizeof(dis)); memset(pre,-1,sizeof(pre)); dis[s]=0; queue<int> q; q.push(s); while(!q.empty()){ int u=q.front(); q.pop(); vis[u]=0; for(int i=head[u];~i;i=Next[i]){ int v=To[i]; int d=ct[i]; if(dis[v]>dis[u]+d&&flow[i]>0){ dis[v]=dis[u]+d; pre[v]=u; id[v]=i; if(!vis[v]){ vis[v]=1; q.push(v); } } } } return dis[t]<inf; } int Maxflow(){ int ans=0; while(spfa()){ int Min=inf; for(int i=t;i!=s;i=pre[i]){ Min=min(flow[id[i]],Min); } for(int i=t;i!=s;i=pre[i]){ flow[id[i]]-=Min; flow[id[i]^1]+=Min; } ans+=dis[t]; } return ans; } int main() { #ifndef ONLINE_JUDGE FIN; #endif while(~scanf("%d %d",&n,&m),n+m){ for(int i=1;i<=n;i++) scanf("%s",str[i]+1); memset(head,-1,sizeof(head)); cnt=1; get_graph(); printf("%d ",Maxflow()); } return 0; }