【POJ 2195】 Going Home（KM算法求最小权匹配）
Going Home
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 20303 | Accepted: 10297 |
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
Source
越来越认为KM算法是种非常奇妙的算法，或者说各种图论算法都好奇妙……在或者说 图是一种奇妙的东西。。。一个题可能有多种算法能够解决
这题之前一仅仅用最小费做的，一直以为是唯一的做法。
今天做KM突然发现 这题之前不是做过么……
这题用KM算法会快非常多。并且代码量少一点。
只是思想的算法。没法做什么比較。
题目大意就是一定数量的人和房子。保证人和房子数量同样，要求分配每一个人到相应的房子里，而且人与房子一一相应。问全部人须要移动的最少步数。建图我是把人和房子的坐标用pair存成数组，然后求出每一个人到每一个房子的距离。也就是最初的二分图。
之后用KM求最小权，跟最大权一种套路，仅仅只是建图的时候权值取负，这样求出的最大权事实上绝对值是最小的。换句话说取反回来后，事实上就是所求的解。
代码例如以下：
#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define fread() freopen("in.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)
#define Pr pair<int,int>
using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const double eps = 1e-8;
//二分图
int mp[233][233];
// x顶标 y顶标
int lx[233],ly[233],link[233],slick[233];
bool visx[233],visy[233];
// 人的坐标 房子坐标
Pr human[233],house[233];
//人数 房数
int nm,nh;
bool cal(int x)
{
visx[x] = 1;
for(int y = 0; y < nh; ++y)
{
if(visy[y]) continue;
int t = lx[x]+ly[y]-mp[x][y];
if(t == 0)
{
visy[y] = 1;
if(link[y] == -1 || cal(link[y]))
{
link[y] = x;
return true;
}
}
else slick[y] = min(slick[y],t);
}
return false;
}
int KM()
{
memset(link,-1,sizeof(link));
for(int i = 0; i < nm; ++i)
{
memset(slick,INF,sizeof(slick));
while(1)
{
memset(visx,0,sizeof(visx));
memset(visy,0,sizeof(visy));
if(cal(i)) break;
int d = INF;
for(int j = 0; j < nh; ++j)
if(!visy[j]) d = min(d,slick[j]);
for(int j = 0; j < nm; ++j)
if(visx[j]) lx[j] -= d;
for(int j = 0; j < nh; ++j)
if(visy[j]) ly[j] += d;
else slick[j] -= d;
}
}
int ans = 0;
for(int i = 0; i < nh; ++i)
if(link[i] != -1) ans += mp[link[i]][i];
return ans;
}
int main()
{
int n,m;
char tmp[233];
while(~scanf("%d%d",&n,&m) && (n+m))
{
nm = nh = 0;
for(int i = 0; i < n; ++i)
{
scanf("%s",tmp);
for(int j = 0; j < m; ++j)
if(tmp[j] == 'm') human[nm++] = Pr(i,j);
else if(tmp[j] == 'H') house[nh++] = Pr(i,j);
}
memset(lx,0,sizeof(lx));
memset(ly,0,sizeof(ly));
for(int i = 0; i < nm; ++i)
for(int j = 0; j < nh; ++j)
{
mp[i][j] = -(abs(human[i].first-house[j].first)+abs(human[i].second-house[j].second));
lx[i] = max(lx[i],mp[i][j]);
}
printf("%d
",-KM());
}
return 0;
}