POJ2195:Going Home
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 26210 | Accepted: 13134 |
题目链接：http://poj.org/problem?id=2195
Description:
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.
Input:
There 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.
Output:
For 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
题意：
给出一个n*m的棋盘，有些点上有房子，有些点上有人，最后人都要进房子里（房子的数量和人的数量想等），每个人每走一步就有1的金币。
问当所有人都进房子里面时，所给出的金币数量最少为多少。
题解：
求出每个人到房子的曼哈顿距离并建图后，直接跑最大流最小费用流就好了。
代码如下：
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <vector> #include <cmath> #include <queue> using namespace std; const int N = 105 ,INF = 99999999,t = 505; int n,m,cnt,tot,num; int map[N][N],d[N][N],head[N*N],dis[N*N],vis[N*N],a[N*N],p[N*N],pre[N*N]; int Dis(int x1,int y1,int x2,int y2){ return abs(x1-x2)+abs(y1-y2); } struct Edge{ int u,v,next,c,w; }e[N*N*N]; void adde(int u,int v,int w){ e[tot].v=v;e[tot].c=1;e[tot].w=w;e[tot].next=head[u];head[u]=tot++; e[tot].v=u;e[tot].c=0;e[tot].w=-w;e[tot].next=head[v];head[v]=tot++; } int BellmanFord(int s,int t,int &flow,int &cost){ for(int i=0;i<=t+5;i++) dis[i]=INF,a[i]=INF; queue <int> q;memset(vis,0,sizeof(vis));memset(p,-1,sizeof(p)); memset(pre,-1,sizeof(pre)); q.push(s);vis[s]=1;dis[s]=0;p[s]=0; while(!q.empty()){ int u=q.front();q.pop();vis[u]=0; for(int i=head[u];i!=-1;i=e[i].next){ int v=e[i].v; if(e[i].c>0 &&dis[v]>dis[u]+e[i].w){ dis[v]=dis[u]+e[i].w; a[v]=min(a[u],e[i].c); p[v]=u; pre[v]=i; if(!vis[v]){ vis[v]=1; q.push(v); } } } } if(dis[t]==INF) return 0; flow+=a[t]; cost+=a[t]*dis[t]; for(int i=t;i>0;i=p[i]){// e[pre[i]].c-=a[t]; e[pre[i]^1].c+=a[t]; } return 1; } int Min_cost(){ int flow=0,cost=0; while(BellmanFord(0,t,flow,cost)); return cost; } int main(){ while(~scanf("%d%d",&n,&m)){ if(n==0 && m==0) break; tot=0;num=0;cnt=0; memset(head,-1,sizeof(head)); vector <pair<int,int> > h; memset(map,0,sizeof(map)); for(int i=1;i<=n;i++){ char s[N]; scanf("%s",s); for(int j=0;j<m;j++){ if(s[j]=='H'){ map[i][j+1]=2; h.push_back(make_pair(i,j+1)); } if(s[j]=='m') map[i][j+1]=1,num++; } } for(int i=1;i<=num;i++) adde(0,i,0); for(int i=num+1;i<=2*num;i++) adde(i,t,0); for(int i=1;i<=n;i++){ for(int j=1;j<=m;j++){ if(map[i][j]!=1) continue ; cnt++;int tmp=0; for(int k=0;k<h.size();k++){ tmp++; pair <int,int> v = h[k]; adde(cnt,num+tmp,Dis(i,j,v.first,v.second)); } } } printf("%d\n",Min_cost()); } return 0; }