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
28
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define N 110
#define INF 99999999
int 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;
}
#include<iostream>
#include<cstring>
#include<climits>
#include<cstdio>
#include<algorithm>
#define N 110
using namespace std;
char maps[N][N]; //存储原始字符地图的
int map[N][N]; //类似边表的可匹配存储
int lx[N], ly[N];
int slack[N];
int match[N];
bool visitx[N], visity[N];
int n;
bool Hungary( int u ) //匹配
{
int i ;
visitx[u] = true;
for( i=0; i < n; ++i)
{
if(visity[i]==true )
continue;
else
{
if(lx[u] + ly[i] == map[u][i] )
{
visity[i] = true;
if(match[i] == -1 || Hungary(match[i]) )
{
match[i] = u;
return true;
}
}
else
slack[i] = min(slack[i], lx[u] + ly[i]-map[u][i] );
}
}
return false;
}
void KM_perfect_match() //匈牙利算法
{
int temp;
memset(lx, 0, sizeof(lx)); // 清零？？
memset(ly, 0, sizeof(ly)); // 清零？？
for(int i=0; i<n; ++i)
for(int j=0; j<n; ++j)
lx[i] = max( lx[i], map[i][j] ); //表示当前的i号人，去某一个房子的最大距离
for(int i=0; i<n; ++i)
{
/*
我们给每个Y顶点一个“松弛量”函数slack，每次开始找增广路时初始化为无穷大。
在寻找增广路的过程中，检查边(i,j)时，如果它不在相等子图中，则让slack[j]变成
原值与A[i]+B[j]-w[i,j]的较小值。
这样，在修改顶标时，取所有不在交错树中的Y顶点的slack值中的最小值作为d值即可。
但还要注意一点：修改顶标后，要把所有的不在交错树中的Y顶点的slack值都减去d。
*/
for(int j=0; j<n; ++j)
slack[j] = INT_MAX;
while(1)
{
memset(visitx, false, sizeof(visitx)); //清零
memset(visity, false, sizeof(visity)); //清零
if( Hungary(i) )
break;
else
{
temp = INT_MAX;
for(int j=0; j<n; ++j )
{
if(!visity[j])
{
temp = min(temp, slack[j]);
}
}
for(int j=0; j<n; ++j )
{
if( visitx[j] )
lx[j] -= temp;
if( visity[j] )
ly[j] += temp;
else
slack[j] -= temp;
}
}
}
}
}
int main()
{
int row, col, ans, numi, numj;
while(scanf("%d %d", &row, &col) && (row + col) ) //行 列
{
n = ans = numi = numj = 0;
memset(match, -1, sizeof(match)); //match数组初始 -1
for(int i=0; i<row; ++i)
{
scanf("%*c");//取回车
for(int j=0; j<col; ++j)
{
scanf("%c", &maps[i][j]);
if(maps[i][j] == 'm')
n++; //记录 人数
}
}
//果然是四层循环啊,和预想的一样
for(int i=0; i<row; ++i) //建图
{
for(int j=0; j<col; ++j)
{ //如果当前的是: 人
if(maps[i][j] == 'm')
{ //暴力一遍整个map,
for(int k=0; k<row; ++k )
{
for(int l = 0; l < col; ++l)
{
if(maps[k][l] == 'H') // 如果当前找到了一所房子
{ //建图时的特殊处理（类似入栈, 不过这次入得是二维数组
map[numi][numj++] = 100 - (abs(k - i) + abs(l - j)); //大数减边
} //等同于给每个人开了一个一位数组
}
}
numi++ ; //当找到下一个人的时候, 二维数组挪到下一行
numj = 0 ; //位置指针归零
}
}
}
KM_perfect_match(); //调用匈牙利算法
for(int i = 0; i < n; ++i )
{
ans = ans + map[match[i]][i];
}
printf("%d\n", 100 * n - ans) ;
}
return 0;
}
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