POJ2195 Going Home 最小費用最大流

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：
對每條邊取相反數，然后得到的結果再取相反數，就能得到最小權匹配

#include <cmath> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define INF 0x3f3f3f3f #define MAXN 105using namespace std;int n;//二分圖頂點集的大小 int a[MAXN][MAXN];//a[i][j]表示i到j的權值 int vx[MAXN],vy[MAXN]; int lx[MAXN],ly[MAXN];//同時調節兩個數組，使得權值和最大 int pre[MAXN];//記錄頂點集n2中點x在頂點集n1中所匹配的點x的編號 int slack[MAXN];//松弛操作 int N,M;//N行M列 int cnth,cntm; struct point{int x,y; }tmph[MAXN],tmpm[MAXN]; char st[MAXN][MAXN];bool dfs(int index) {vx[index]=1;for(int i=1;i<=n;i++){if(!vy[i]&&lx[index]+ly[i]==a[index][i]){//這個條件下使用匈牙利算法vy[i]=1;//這個房子沒人住或者可以讓住這個房子的人找另外的房子if(!pre[i]||dfs(pre[i])){pre[i]=index;return 1;}}else if(!vy[i]&&lx[index]+ly[i]>a[index][i])slack[i]=min(slack[i], lx[index]+ly[i]-a[index][i]);}return 0; }int KM() {memset(ly,0,sizeof(ly));//首先把每個居民出錢最多的那個房子給他for(int i=1;i<=n;i++){lx[i]=-INF;for(int j=1;j<=n;j++)lx[i]=max(lx[i],a[i][j]);}int ans=0,d;memset(pre,0,sizeof(pre));//給第i位居民分配房子for(int i=1;i<=n;i++){memset(slack,INF,sizeof(slack));while(1){memset(vx,0,sizeof(vx));memset(vy,0,sizeof(vy));if(dfs(i))break;d=INF;for(int i=1;i<=n;i++)if(!vy[i])d=min(d,slack[i]);//找到最小松弛量if(d==INF) return -1; //no matchingfor(int j=1;j<=n;j++){if(vx[j]) lx[j]-=d;if(vy[j]) ly[j]+=d;}}}for(int i=1;i<=n;i++)ans+=a[pre[i]][i];return ans; }int main() {while(~scanf("%d %d", &N, &M)){if(!N&&!M)break;cnth=cntm=0;getchar();for(int i=1;i<=N;i++){for(int j=1;j<=M;j++){scanf("%c",&st[i][j]);if(st[i][j]=='m')cntm++,tmpm[cntm].x=i, tmpm[cntm].y=j;if(st[i][j]=='H')cnth++,tmph[cnth].x=i, tmph[cnth].y=j;}getchar();}n=cntm;for(int i=1;i<=cntm;i++)for(int j=1;j<=cnth;j++)a[i][j]=-(abs(tmpm[i].x-tmph[j].x)+abs(tmpm[i].y-tmph[j].y));printf("%d\n",-KM());}return 0; }

Java

import java.math.*; import java.util.*;public class Main {static int inf=(int)1e8+7;static int maxn=110;static int n;//n個球，n個洞static int[][] w=new int[maxn][maxn];//w[i][j]表示i到j的權值static int[] lx=new int[maxn];static int[] ly=new int[maxn];static int[] link=new int[maxn];static int[] slack=new int[maxn];static boolean[] visx=new boolean[maxn];static boolean[] visy=new boolean[maxn];static boolean dfs(int x) {visx[x]=true;for(int y=1;y<=n;y++) {//if(visy[y]) continue;int t=lx[x]+ly[y]-w[x][y];if(!visy[y]&&t==0) {visy[y]=true;if(link[y]==0||dfs(link[y])) {link[y]=x;return true;}}else if(!visy[y]&&lx[x]+ly[y]>w[x][y])slack[y]=Math.min(slack[y], lx[x]+ly[y]-w[x][y]);}return false;}static int km() {for(int i=0;i<maxn;i++) {lx[i]=-inf;ly[i]=0;link[i]=0;}for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)lx[i]=Math.max(lx[i], w[i][j]);for(int i=1;i<=n;i++) {for(int j=1;j<=n;j++) slack[j]=inf;while(true) {for(int k=1;k<maxn;k++) {visx[k]=false;visy[k]=false;}if(dfs(i)) break;int d=inf;for(int k=1;k<=n;k++)if(!visy[k]&&d>slack[k])d=slack[k];if(d==-1) return -1;// no matchingfor(int k=1;k<=n;k++) {if(visx[k]) lx[k]-=d;if(visy[k]) ly[k]+=d;}}}int ans=0;for(int i=1;i<=n;i++)ans+=w[link[i]][i];return ans;}public static void main (String[] args) {Scanner cin=new Scanner(System.in);while(cin.hasNext()) {int[] ball_x=new int[maxn];int[] ball_y=new int[maxn];int[] hole_x=new int[maxn];int[] hole_y=new int[maxn];int bk=1,hk=1;int M=cin.nextInt();int N=cin.nextInt();String str="";if(M==0&&N==0) break;for(int i=0;i<M;i++) {str=cin.next();for(int j=0;j<N;j++) {if(str.charAt(j)=='m') {ball_x[bk]=i;ball_y[bk]=j;bk++;}else if(str.charAt(j)=='H') {hole_x[hk]=i;hole_y[hk]=j;hk++;}}}n=bk-1;//System.out.println(n);for(int i=1;i<=n;i++) {for(int j=1;j<=n;j++) {w[i][j]=-Math.abs(ball_x[i]-hole_x[j])-Math.abs(ball_y[i]-hole_y[j]);//System.out.print(-w[i][j]+" ");}//System.out.println();}System.out.println(-km());}cin.close();} }

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