HDU 3435 KM A new Graph Game

和HDU 3488一样的，只不过要判断一下是否有解。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 using namespace std;

 const int maxn =  + ;
 const int INF = 0x3f3f3f3f;

 int n, m;

 int W[maxn][maxn], lft[maxn];
 int slack[maxn];
 int Lx[maxn], Ly[maxn];
 bool S[maxn], T[maxn];

 bool match(int u)
 {
     S[u] = true;
     for(int v = ; v <= n; v++) if(!T[v])
     {
         int t = Lx[u] + Ly[v] - W[u][v];
         if( == t)
         {
             T[v] = true;
             if(!lft[v] || match(lft[v]))
             {
                 lft[v] = u;
                 return true;
             }
         }
         else slack[v] = min(slack[v], t);
     }

     return false;
 }

 void update()
 {
     int a = INF;
     for(int i = ; i <= n; i++) if(!T[i]) a =  min(a, slack[i]);
     for(int i = ; i <= n; i++)
     {
         if(S[i]) Lx[i] -= a;

         if(T[i]) Ly[i] += a;
         else  slack[i] -= a;
     }
 }

 void KM()
 {
     memset(Ly, , sizeof(Ly));
     memset(lft, , sizeof(lft));
     for(int i = ; i <= n; i++) Lx[i] = -INF;
     for(int i = ; i <= n; i++)
         for(int j = ; j <= n; j++) Lx[i] = max(Lx[i], W[i][j]);

     for(int i = ; i <= n; i++)
     {
         memset(slack, 0x3f, sizeof(slack));
         for(;;)
         {
             memset(S, false, sizeof(S));
             memset(T, false, sizeof(T));
             if(match(i)) break;
             update();
         }
     }
 }

 int main()
 {
     int T; scanf("%d", &T);
     for(int kase = ; kase <= T; kase++)
     {
         scanf("%d%d", &n, &m);
         for(int i = ; i <= n; i++)
             for(int j = ; j <= n; j++)
                 W[i][j] = -INF;
         for(int i = ; i < m; i++)
         {
             int u, v, d; scanf("%d%d%d", &u, &v, &d);
             if(W[u][v] < -d) W[u][v] = W[v][u] = -d;
         }

         KM();

         bool ok = true;
         int ans = ;
         for(int i = ; i <= n; i++)
         {
             if(W[lft[i]][i] == -INF) { ok = false; break; }
             ans += W[lft[i]][i];
         }
         printf("Case %d: ", kase);
         if(!ok) puts("NO");
         else printf("%d\n", -ans);
     }

     return ;
 }

代码君