HDU 3315 KM My Brute

参考题解

二分图的最优匹配。图很容易建立。再处理相似度的时候。把每个权值扩大100倍。然后再对i==j时 特殊标记。使他们的权值再++1。后面选择的时候就很容易挑出。按原匹配

匹配的个数。 100*（double）(res%100)/n。即可得到第二问。

 

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <algorithm>
  5 #include <vector>
  6 #include <queue>
  7 using namespace std;
  8 
  9 const int maxn = 100;
 10 const int INF = 0x3f3f3f3f;
 11 
 12 int n;
 13 
 14 int V[maxn], H[maxn], P[maxn], A[maxn], B[maxn];
 15 
 16 // KM algorithm
 17 int W[maxn][maxn], lft[maxn];
 18 int slack[maxn];
 19 int Lx[maxn], Ly[maxn];
 20 bool S[maxn], T[maxn];
 21 
 22 int inline divide(int a, int b) { int ans = a / b; if(a % b) ans++; return ans; }
 23 
 24 bool match(int u)
 25 {
 26     S[u] = true;
 27     for(int v = 1; v <= n; v++) if(!T[v])
 28     {
 29         int t = Lx[u] + Ly[v] - W[u][v];
 30         if(t == 0)
 31         {
 32             T[v] = true;
 33             if(!lft[v] || match(lft[v]))
 34             {
 35                 lft[v] = u;
 36                 return true;
 37             }
 38         }
 39         else slack[v] = min(slack[v], t);
 40     }
 41 
 42     return false;
 43 }
 44 
 45 void update()
 46 {
 47     int a = INF;
 48     for(int i = 1; i <= n; i++) if(!T[i]) a = min(a, slack[i]);
 49     for(int i = 1; i <= n; i++)
 50     {
 51         if(S[i]) Lx[i] -= a;
 52 
 53         if(T[i]) Ly[i] += a;
 54         else slack[i] -= a;
 55     }
 56 }
 57 
 58 void KM()
 59 {
 60     memset(Ly, 0, sizeof(Ly));
 61     memset(lft, 0, sizeof(lft));
 62     for(int i = 1; i <= n; i++) Lx[i] = -INF;
 63     for(int i = 1; i <= n; i++)
 64         for(int j = 1; j <= n; j++) Lx[i] = max(Lx[i], W[i][j]);
 65 
 66     for(int i = 1; i <= n; i++)
 67     {
 68         memset(slack, 0x3f, sizeof(slack));
 69         for(;;)
 70         {
 71             memset(S, false, sizeof(S));
 72             memset(T, false, sizeof(T));
 73             if(match(i)) break;
 74             update();
 75         }
 76     }
 77 }
 78 
 79 int main()
 80 {
 81     while(scanf("%d", &n) == 1 && n)
 82     {
 83         for(int i = 1; i <= n; i++) scanf("%d", V + i);
 84         for(int i = 1; i <= n; i++) scanf("%d", H + i);
 85         for(int i = 1; i <= n; i++) scanf("%d", P + i);
 86         for(int i = 1; i <= n; i++) scanf("%d", A + i);
 87         for(int i = 1; i <= n; i++) scanf("%d", B + i);
 88 
 89         //build graph
 90         for(int i = 1; i <= n; i++)
 91             for(int j = 1; j <= n; j++)
 92             {
 93                 int score = V[i] * 100;
 94                 int round1 = divide(P[j], A[i]);
 95                 int round2 = divide(H[i], B[j]);
 96                 if(round1 > round2) score = -score;
 97                 if(i == j) score++;
 98                 W[i][j] = score;
 99             }
100 
101         KM();
102 
103         int ans = 0;
104         for(int i = 1; i <= n; i++) ans += W[lft[i]][i];
105 
106         if(ans < 0)
107         {
108             puts("Oh, I lose my dear seaco!");
109             continue;
110         }
111 
112         printf("%d ", ans / 100);
113         printf("%.3f%%\n", 100.0 * (ans % 100) / n);
114     }
115 
116     return 0;
117 }
代码君

 

posted @ 2015-09-11 20:29  AOQNRMGYXLMV  阅读(244)  评论(0)  编辑  收藏  举报