[JSOI2008]巨额奖金（最小生成树计数）

基于kruscal的贪心 可以有如下推论：两个不同的最小生成树所用的的相同边权的边的数量相同。

可以做这样的理解：当进行到选第K大的边时，联通快数是一定的所以的、一定需要K- 1个该权值边 （不知道对不对 欢迎TuneProverbs指正）

SB没想好怎么存水体写半天。。。。。。。

我是SB 
  1 #include <cstdio>
  2 #include <vector>
  3 #include <cstring>
  4 #include <iostream>
  5 #include <algorithm>
  6 using namespace std;
  7 const int N = 1500;typedef long long LL;
  8 vector<int> v[N];
  9 vector<int>::iterator it, ptr;
 10 int n, m, p[N], cnt[N], cc, r[N];
 11 LL ans;
 12 bool vis[N], cantvisit[N], noans;
 13 struct Q {
 14     int x, y, w, num;
 15     inline bool operator < (const Q & a) const {
 16         return w < a.w;
 17     }
 18 }e[N];
 19 inline int F (int x) { return p[x] == x ? x : p[x] = F (p[x]); }
 20 void ksc ()
 21 {
 22     for (int i = 1; i <= n; i ++)
 23         p[i] = i;
 24     sort (e + 1, e + 1 + m);
 25     for (int i = 1; i <= m; i ++)
 26         if (F (e[i].x) != F (e[i].y))
 27         {
 28             p[F (e[i].x)] = F (e[i].y);
 29             ans += e[i].w;
 30             int j (0);
 31             for (int j = 1; j <= cc; j ++)
 32                 if (r[j] == e[i].w)
 33                     cnt[j] ++;
 34         }
 35     for (int i = 2; i <= n; i ++)
 36         if (F (i) != F (i - 1))
 37             noans = true;
 38     //for (int i = 1; i <= 5; i ++)
 39     //    printf ("%d\n", cnt[i]);
 40     //printf ("%d\n", ans);
 41 }
 42 bool ck ()
 43 {
 44     for (int i = 1; i <= n; i ++)
 45         p[i] = i;
 46     sort (e + 1, e + 1 + m);
 47     LL z (0);
 48     for (int i = 1; i <= m; i ++)
 49     {
 50         if (cantvisit[e[i].num]) continue;
 51         if (F (e[i].x) != F (e[i].y))
 52         {
 53             p[F (e[i].x)] = F (e[i].y);
 54             z += e[i].w;
 55         }
 56     }
 57     //printf ("%d\n", z);
 58     return z == ans;
 59 }
 60 inline int calc (int j)
 61 {
 62     int z (0);
 63     for (; j; j -= j & -j) z ++;
 64     return z;
 65 }
 66 int main ()
 67 {
 68     scanf ("%d%d", &n, &m);
 69     for (int i = 1; i <= m; i ++)
 70         scanf ("%d%d%d", &e[i].x, &e[i].y, &e[i].w), e[i].num = i;
 71     for (int i = 1; i <= m; i ++)
 72     {
 73         if (vis[i]) continue;
 74         cc ++;
 75         r[cc] = e[i].w;
 76         v[cc].push_back (i);
 77         for (int j = i + 1; j <= m; j ++)
 78             if (e[j].w == e[i].w)
 79             {
 80                 vis[j] = true;
 81                 v[cc].push_back (j);
 82             }
 83     }
 84     ksc ();
 85     if (noans == true)
 86     {
 87         puts ("0");
 88         return 0;
 89     }
 90     int z (1);
 91     for (int j = 1; j <= cc; j ++)
 92     {
 93         int tmp (0);
 94         //printf ("*%d\n", v[j].size ());
 95         for (int i = 0; i < (1 << v[j].size ()); i ++)
 96         {
 97             if (calc (i) == cnt[j])
 98             {
 99                 int k (0);
100                 for (ptr = v[j].begin (); ptr != v[j].end (); ptr ++, k ++)
101                     if ((i >> k) & 1)
102                         ;
103                     else
104                         cantvisit[*ptr] = true;
105                 //for (int i = 1; i <= m; i ++)
106                 //    printf ("%d", cantvisit[i]);puts ("");
107                 int t (0);
108                 if (t = ck ())
109                     tmp ++;
110                 memset (cantvisit, 0, sizeof cantvisit);
111             }
112         }
113         //printf ("%d\n", tmp);
114         if (tmp != 0)
115             z = ((long long)z * tmp) % 31011;
116     }
117     printf ("%d\n", z);
118     return 0;
119 }

 

posted @ 2013-02-04 10:55  Moretimes  阅读(330)  评论(0)  编辑  收藏  举报