bzoj1004 [HNOI2008]Cards

传送门：https://www.lydsy.com/JudgeOnline/problem.php?id=1004

【题解】

Burnside引理，考虑Polya原理的推导，由于循环节必须染相同的颜色，那么可以dp出方案。

 1 # include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int N = 110;
 5 int A, B, C, m, mod, n;
 6 int p[N][N];
 7 bool vis[N];
 8 int w[N], wn;
 9 int f[N][N][N];
10 
11 inline int dp(int id) {
12     wn = 0;
13     for (int i=1; i<=n; ++i) vis[i] = 0;
14     for (int i=1; i<=n; ++i) {
15         if(vis[i]) continue;
16         int x = i, t = 0;
17         while(vis[x] == 0) {
18             vis[x] = 1; 
19             x = p[id][x];
20             ++ t;
21         }
22         w[++wn] = t;
23     }
24     for (int i=0; i<=A; ++i)
25         for (int j=0; j<=B; ++j)
26             for (int k=0; k<=C; ++k) f[i][j][k] = 0;
27     f[0][0][0] = 1;
28     for (int cur=1; cur<=wn; ++cur)
29         for (int i=A; i>=0; --i)
30             for (int j=B; j>=0; --j)
31                 for (int k=C; k>=0; --k) {
32                     if(i >= w[cur]) {
33                         f[i][j][k] = f[i][j][k] + f[i-w[cur]][j][k];
34                         if(f[i][j][k] >= mod) f[i][j][k] -= mod;
35                     }
36                     if(j >= w[cur]) {
37                         f[i][j][k] = f[i][j][k] + f[i][j-w[cur]][k];
38                         if(f[i][j][k] >= mod) f[i][j][k] -= mod;
39                     }
40                     if(k >= w[cur]) {
41                         f[i][j][k] = f[i][j][k] + f[i][j][k-w[cur]];
42                         if(f[i][j][k] >= mod) f[i][j][k] -= mod;
43                     }
44                 }
45     return f[A][B][C];
46 }
47 
48 inline int pwr(int a, int b) {
49     int ret = 1;
50     while(b) {
51         if(b&1) ret = ret * a % mod;
52         a = a * a % mod;
53         b >>= 1;
54     }
55     return ret;
56 }
57     
58         
59 int main() {
60     cin >> A >> B >> C >> m >> mod;
61     n = A + B + C;
62     for (int i=1; i<=m; ++i) 
63         for (int j=1; j<=n; ++j) cin >> p[i][j];
64     ++m;
65     for (int i=1; i<=n; ++i) p[m][i] = i;
66     int ans = 0;
67     for (int i=1; i<=m; ++i) {
68         ans = ans + dp(i);
69         if(ans >= mod) ans -= mod;
70     }
71     ans = ans * pwr(m, mod-2) % mod;
72     cout << ans << endl;
73     return 0;
74 }
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posted @ 2019-02-10 16:12  Galaxies  阅读(240)  评论(0)  编辑  收藏  举报