HDU 1533 Going Home （二分图最小权匹配 KM模板）

题意：n*m的方格中有p个房子和p个人，人走一步花费1元，要求每个人都要找到一个房子，且每个房子里只能有一个人，问p个人都找到各自的房子，最小的花费是多少。

思路：以每个人到所有房子的曼哈顿距离的相反数为权值建边，套下KM就是最小权匹配。

#include
#include
#include
#include
#include
using namespace std;
const int maxn = 505;
const int INF = 0x3f3f3f3f;
struct node
{
    int x, y;
    node() {}
    node(int xx, int yy): x(xx), y(yy) {}
}a[maxn], b[maxn];
//
int nx, ny;
int match[maxn], lx[maxn], ly[maxn], slack[maxn];
int visx[maxn], visy[maxn], w[maxn][maxn];

int Hungary(int x)
{
    visx[x] = 1;
    for(int y = 1; y <= ny; y++)
    {
        if(visy[y]) continue;
        int temp = lx[x]+ly[y]-w[x][y];
        if(temp == 0)
        {
            visy[y] = 1;
            if(match[y] == -1 || Hungary(match[y]))
            {
                match[y] = x;
                return 1;
            }
        }
        else if(slack[y] > temp)
            slack[y] = temp;
    }
    return 0;
}

int KM()
{
    memset(match, -1, sizeof(match));
    memset(ly, 0, sizeof(ly));
    for(int i = 1; i <= nx; i++)
        lx[i] = -INF;
    for(int i = 1; i <= nx; i++)
        for(int j = 1; j <= ny; j++)
            if(w[i][j] > lx[i])
                lx[i] = w[i][j];
    for(int x = 1; x <= nx; x++)
    {
        for(int i = 1; i <= ny; i++)
            slack[i] = INF;
        while(1)
        {
            memset(visx, 0, sizeof(visx));
            memset(visy, 0, sizeof(visy));
            if(Hungary(x)) break;
            int d = INF;
            for(int i = 1; i <= ny; i++)
                if(!visy[i] && d > slack[i])
                    d = slack[i];
            for(int i = 1; i <= nx; i++)
                if(visx[i])
                    lx[i] -= d;
            for(int i = 1; i <= ny; i++)
                if(visy[i])
                    ly[i] += d;
                else
                    slack[i] -= d;
        }
    }
    int res = 0;
    for(int i = 1; i <= ny; i++)
        if(match[i] != -1)
            res += w[match[i]][i];
    return res;
}

int main(void)
{
    int n, m;
    while(cin >> n >> m && n+m)
    {
        char ch;
        int cnt1 = 1, cnt2 = 1;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
            {
                scanf(" %c", &ch);
                if(ch == 'm') a[cnt1++] = node(i, j);
                if(ch == 'H') b[cnt2++] = node(i, j);
            }
        nx = ny = cnt1-1;
        for(int i = 1; i <= nx; i++)
            for(int j = 1; j <= nx; j++)
                w[i][j] = -abs(a[i].x-b[j].x)-abs(a[i].y-b[j].y);
        printf("%d\n", -KM());
    }
    return 0;
}

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