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HDU 1533 Going home
Going Home
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2829 Accepted Submission(s): 1423
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.
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
#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#define N 110#define INF 99999999int n, m;char map[N][N]; //存储原始字符地图的int ma[N][N]; //类似边表的可匹配存储int lx[N], ly[N];int vtx[N], vty[N];int match[N];int slack[N];int cnt;int max(int a, int b){ return a>b?a:b;}int min(int a, int b){ return a>b?b:a;}int hungary(int dd) //匈牙利算法{ int i; vtx[dd]=1; for(i=0; i<cnt; i++) { if(vty[i]) continue; else { if(lx[dd]+ly[i] == ma[dd][i] ) { vty[i]=1; if(match[i]==-1 || hungary(match[i]) ) { match[i] = dd; return 1; } } else slack[i] = min( slack[i], lx[dd] + ly[i]-ma[dd][i] ); } } return 0;}void km_match() //最大权匹配{ int i, j; int temp; memset(lx, 0, sizeof(lx)); memset(ly, 0, sizeof(ly)); for(i=0; i<cnt; i++) { for(j=0; j<cnt; j++) { lx[i]=max(lx[i], ma[i][j] ); } //表示当前的i号人，去某一个房子的最大距离 } for(i=0; i<cnt; i++) { for(j=0; j<cnt; j++) { slack[j]=INF; //初始无穷大 } while(1) { memset(vtx, 0, sizeof(vtx)); memset(vty, 0, sizeof(vty)); if(hungary(i)) //匈牙利算法 break; else { temp=INF; for(j=0; j<cnt; j++) { if(!vty[j]) { temp=min(temp, slack[j] ); } } for( j=0; j<cnt; j++ ) { if( vtx[j] ) lx[j] -= temp; if( vty[j] ) ly[j] += temp; else slack[j] -= temp; } } } }}int main(){ int i, j, k, ll; int ci, cj; int sum; while(scanf("%d %d", &n, &m) && n!=0 && m!=0 ) { memset(match, -1, sizeof(match ));//match数组初始 -1,记录父节点 cnt=0; for(i=0; i<n; i++ ) { scanf("%*c"); //每行先取一个回车换行 for(j=0; j<m; j++) { scanf("%c", & map[i][j] ); if(map[i][j] == ‘m‘ ) //如果是个人 { cnt++; //记录 人数, 建图时需要 } } } //四层循环 前两层遍历map寻找m 内两层循环找h ci=0; cj=0; for(i=0; i<n; i++) { for(j=0; j<m; j++) { if(map[i][j]==‘m‘) //找到一个人 { //找到人之后遍历map找 H for(k=0; k<n; k++) { for(ll=0; ll<m; ll++) { if(map[k][ll]==‘H‘) { ma[ci][cj++] = 100-(abs(k-i)+abs(ll-j)); //大数减边 } } } ci++; //换到下一行存储 cj=0; //cj指针回到0位置 } } } km_match(); //最大权匹配 sum=0; for(i=0; i<cnt; i++) { sum+=ma[match[i]][i] ; } printf("%d\n", 100*cnt-sum ); } return 0;}28
Source
Pacific Northwest 2004
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