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 =  + ;

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

 void KM()
 {
     for(int i = ; i <= n; i++)
     {
         Lx[i] = *max_element(W[i]+, W[i]++n);
         Ly[i] = ;
         lft[i] = ;
     }
     for(int u = ; 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) ==  && row)
     {
         n = ;

         for(int i = ; i < row; i++) scanf("%s", grid[i]);
         int hcnt = , mcnt = ;
         for(int i = ; i < row; i++)
             for(int j = ; 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 = ; i <= n; i++)
             for(int j = ; j <= n; j++)
                 W[i][j] = -dist(men[i], house[j]);

         KM();

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

     return ;
 }

代码君