HDU 1533 二分图最小权匹配 Going Home

带权二分图匹配，把距离当做权值，因为是最小匹配，所以把距离的相反数当做权值求最大匹配。

最后再把答案取一下反即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
#include <cmath>
#define MP make_pair
using namespace std;

typedef pair<int, int> PII;

const int maxn = 100 + 10;

int n;
PII house[maxn], men[maxn];
char grid[maxn][maxn];

int row, col;

int dist(const PII& a, const PII& b)
{ return abs(a.first - b.first) + abs(a.second - b.second); }

//KM algorithm
int W[maxn][maxn];
int Lx[maxn], Ly[maxn];
int lft[maxn];
bool S[maxn], T[maxn];

bool match(int u)
{
    S[u] = true;
    for(int v = 1; v <= n; v++)
        if(Lx[u] + Ly[v] == W[u][v] && !T[v])
        {
            T[v] = true;
            if(!lft[v] || match(lft[v]))
            {
                lft[v] = u;
                return true;
            }
        }
    return false;
}

void update()
{
    int a = 100000000;
    for(int u = 1; u <= n; u++) if(S[u])
        for(int v = 1; v <= n; v++) if(!T[v])
            a = min(a, Lx[u] + Ly[v] - W[u][v]);
    for(int u = 1; u <= n; u++)
    {
         if(S[u]) Lx[u] -= a;
         if(T[u]) Ly[u] += a;
    }
}

void KM()
{
    for(int i = 1; i <= n; i++)
    {
        Lx[i] = *max_element(W[i]+1, W[i]+1+n);
        Ly[i] = 0;
        lft[i] = 0;
    }
    for(int u = 1; u <= n; u++)
    {
        for(;;)
        {
            memset(S, false, sizeof(S));
            memset(T, false, sizeof(T));
            if(match(u)) break;
            else update();
        }
    }
}

int main()
{
    while(scanf("%d%d", &row, &col) == 2 && row)
    {
        n = 0;

        for(int i = 0; i < row; i++) scanf("%s", grid[i]);
        int hcnt = 0, mcnt = 0;
        for(int i = 0; i < row; i++)
            for(int j = 0; j < col; j++)
            {
                if(grid[i][j] == 'H')
                    house[++hcnt] = MP(i, j);
                else if(grid[i][j] == 'm')
                    men[++mcnt] = MP(i, j);
            }

        //build graph
        n = hcnt;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                W[i][j] = -dist(men[i], house[j]);

        KM();

        int ans = 0;
        for(int i = 1; i <= n; i++) ans += W[lft[i]][i];
        printf("%d
", -ans);
    }

    return 0;
}