UVA11383 Golden Tiger Claw —— KM算法

题目链接：https://vjudge.net/problem/UVA-11383

题解：

根据KM()算法，标杆满足：l(x) + l(y) >= w(x, y) 。

当求完最大权匹配之后，所有标杆在满足：l(x) + l(y) >= w(x, y) 的条件下，和最小。

代码如下：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int mod = 1e9+;
 const int MAXN = +;
 
 int nx, ny;
 int g[MAXN][MAXN], linker[MAXN], lx[MAXN], ly[MAXN], slack[MAXN];
 bool visx[MAXN], visy[MAXN];
 
 bool DFS(int x)
 {
     visx[x] = true;
     for(int y = ; y<=ny; y++)
     {
         if(visy[y]) continue;
         int tmp = lx[x] + ly[y] - g[x][y];
         if(tmp==)
         {
             visy[y] = true;
             if(linker[y]==- || DFS(linker[y]))
             {
                 linker[y] = x;
                 return true;
             }
         }
         else
             slack[y] = min(slack[y], tmp);
     }
     return false;
 }
 
 void KM()
 {
     memset(linker, -, sizeof(linker));
     memset(ly, , sizeof(ly));
     for(int i = ; i<=nx; i++)
     {
         lx[i] = -INF;
         for(int j = ; j<=ny; j++)
             lx[i] = max(lx[i], g[i][j]);
     }
 
     for(int x = ; x<=nx; x++)
     {
         for(int i = ; i<=ny; i++)
             slack[i] = INF;
         while(true)
         {
             memset(visx, , sizeof(visx));
             memset(visy, , sizeof(visy));
 
             if(DFS(x)) break;
             int d = INF;
             for(int i = ; i<=ny; i++)
                 if(!visy[i])
                     d = min(d, slack[i]);
 
             for(int i = ; i<=nx; i++)
                 if(visx[i])
                     lx[i] -= d;
             for(int i = ; i<=ny; i++)
             {
                 if(visy[i]) ly[i] += d;
                 else slack[i] -= d;
             }
         }
     }
 }
 
 int main()
 {
     int n;
     while(scanf("%d",&n)!=EOF)
     {
         nx = ny = n;
         for(int i = ; i<=nx; i++)
         for(int j = ; j<=ny; j++)
             scanf("%d",&g[i][j]);
 
         KM();
         int sum = ;
         for(int i = ; i<=nx; i++) printf("%d ", lx[i]), sum += lx[i];
         printf("\n");
         for(int i = ; i<=ny; i++) printf("%d ", ly[i]), sum += ly[i];
         printf("\n");
         printf("%d\n", sum);
     }
 }