POJ 2195 Going Home 二分圖的最大權匹配

Going Home

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?25567	?	Accepted:?12838

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

Source

Pacific Northwest 2004

題意：

有k個人，k間房子，每人進入一個房子，求最小的總距離
分析：

關鍵在于建圖：

源點為0，匯點為2*n+1,n為房子個數（房子數等于人數）；

源點到每個人建立一條容量為1，費用為0的邊；

每個人到每個房子建立一條容量為1，費用為兩者距離的邊；

每個人到匯點建立一條容量為1，費用為0的邊；

然后跑一遍最小費用最大流即可

#include<stdio.h> #include<iostream> #include<string.h> #include<queue> #include<cstdio> #include<string> #include<math.h> #include<algorithm> #include<map> #include<set> #include<stack> #define mod 998244353 #define INF 0x3f3f3f3f #define eps 1e-6 using namespace std; typedef long long ll; #define MAXN 1003 #define MAXM 40004 //最小費用最大流 struct Edge{int to,next;int flow,cost,cap; }edge[MAXM]; int tol,head[MAXN]; void init() {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 inq[MAXN];//標記是否點是否在隊列 int dis[MAXN];//最短距離 int pre[MAXN];//記錄路徑 int q[MAXN*10];//隊列 //單位費用可能是負值，所以用SPFA bool spfa(int st,int en) {memset(inq,0,sizeof inq);memset(dis,INF,sizeof dis);memset(pre,-1,sizeof pre);int rear=0,front=0;dis[st]=0;inq[st]=true;q[front++]=st;while(rear<front){int u=q[rear++];inq[u]=false;for(int e=head[u];e!=-1;e=edge[e].next){int v=edge[e].to;if(edge[e].cap>edge[e].flow&&dis[v]>dis[u]+edge[e].cost){dis[v]=dis[u]+edge[e].cost;pre[v]=e;//表示邊e-->v,e就是v的前驅if(!inq[v])inq[v]=true,q[front++]=v;}}}return pre[en]!=-1; } int minCostMaxFlow(int st,int en,int &cost,int &flow) {//如果能找到從源點到匯點的最短路，說明還沒有達到最小費用最大流while(spfa(st,en)){int Min=INF;//最小殘余流量//沿著當前路徑返回for(int i=pre[en];i!=-1;i=pre[edge[i^1].to]){int rem=edge[i].cap-edge[i].flow;Min=Min>rem?rem:Min;}for(int i=pre[en];i!=-1;i=pre[edge[i^1].to]){edge[i].flow+=Min;//正向邊添加殘余流量edge[i^1].flow-=Min;//反向邊減少殘余流量cost+=Min*edge[i].cost;}flow+=Min;} } //以上為最小費用最大流模板int n,m; struct point{int x,y; }house[MAXN],men[MAXN]; char str[110]; int main() {while(scanf("%d%d",&n,&m)&&n&&m){init();int houseNum=1,menNum=1;for(int i=0;i<n;i++){scanf("%s",str);for(int j=0;j<m;j++){if(str[j]=='H')house[houseNum].x=i,house[houseNum].y=j,houseNum++;if(str[j]=='m')men[menNum].x=i,men[menNum].y=j,menNum++;}}int st=0;int en=houseNum+menNum-1;for(int i=1;i<menNum;i++){for(int j=1;j<houseNum;j++){int w=abs(men[i].x-house[j].x)+abs(men[i].y-house[j].y);addEdge(i,menNum+j-1,1,w);}}for(int i=1;i<menNum;i++){addEdge(0,i,1,0);addEdge(menNum+i-1,en,1,0);}int cost=0,flow=0;minCostMaxFlow(st,en,cost,flow);printf("%d\n",cost);} }

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