Going HomeTime Limit: 10000/5000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3229Accepted Submission(s): 1639
Problem 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.mH.5 5HH..m……………mm..H7 8…H…….H…….H….mmmHmmmm…H…….H…….H….0 0
Sample Output
21028
Source
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题意：二维地图上有人和房子，每个人要到一个房子里去，一个房子只能容纳一个人，每个人的费用为走的距离。求最小费用。
代码：
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6#define lson rt<<1,l,mid#define rson rt<<1|1,mid+1,rtypedef long long ll;using namespace std;/*最小费用最大流，，求最大费用只需要取相反数，结果取相反数即可。点的总数为N，点的编号0~N-1*/const int MAXN = 300;const int MAXM = 100000;struct Node{int x,y;}H[MAXN],M[MAXN];struct Edge{int to,next,cap,flow,cost;}edge[MAXM];int head[MAXN],tol;int pre[MAXN],dis[MAXN];bool vis[MAXN];int N,n,m;void init(int n){N=n;tol=0;memset(head,-1,sizeof(head));}void addedge(int u,int v,int cap,int cost){edge[tol].to=v;edge[tol].cap=cap;edge[tol].cost=cost;edge[tol].flow=0;edge[tol].next=head[u];head[u]=tol++;edge[tol].to=u;edge[tol].cap=0;edge[tol].cost=-cost;edge[tol].flow=0;edge[tol].next=head[v];head[v]=tol++;}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);}}}}if (pre[t]==-1) return false;else return true;}//返回的是最大流，cost存的是最小费用int minCostMaxflow(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;}int main(){char ch;while (scanf("%d%d",&n,&m)&&(m||n)){int hh=1,mm=1;getchar();for (int i=1;i<=n;i++){for (int j=1;j<=m;j++){scanf("%c",&ch);if (ch=='H'){H[hh].x=i;H[hh++].y=j;}else if (ch=='m'){M[mm].x=i;M[mm++].y=j;}}getchar();}N=hh+mm+2;init(N);mm–,hh–;for (int i=1;i<=mm;i++)addedge(0,i,1,0);for (int i=1;i<=hh;i++)addedge(i+mm,N-1,1,0);for (int i=1;i<=mm;i++){for (int j=1;j<=hh;j++){int x=abs(M[i].x-H[j].x)+abs(M[i].y-H[j].y);addedge(i,j+mm,1,x);}}int ans;int y=minCostMaxflow(0,N-1,ans);printf("%d\n",ans);}return 0;}
使你疲倦的不是前面的高山，而是你鞋里的一粒沙子。
