bzoj 1079: [SCOI2008]着色方案

思路:按个数分类dp

 1 #include<bits/stdc++.h>
 2 #define LL long long
 3 #define fi first
 4 #define se second
 5 #define mk make_pair
 6 #define pii pair<int,int>
 7 #define piii pair<int, pair<int,int> >
 8 
 9 using namespace std;
10 
11 const int N = 16;
12 const int M = 10 + 7;
13 const int inf = 0x3f3f3f3f;
14 const LL INF = 0x3f3f3f3f3f3f3f3f;
15 const int mod = 1e9 + 7;
16 
17 int n, cur, sum, num[6];
18 int dp[2][6][N][N][N][N][N];
19 
20 void add(int &a, int b) {
21     a += b; if(a >= mod) a-= mod;
22 }
23 int main() {
24     scanf("%d", &n);
25     for(int i = 1; i <= n; i++) {
26         int x; scanf("%d", &x);
27         num[x]++;
28         sum += x;
29     }
30     dp[cur][0][num[1]][num[2]][num[3]][num[4]][num[5]] = 1;
31 
32     for(int i = 0; i < sum; i++) {
33         cur ^= 1;
34         for(int j = 0; j <= 5; j++)
35             for(int a = 0; a <= n; a++)
36                 for(int b = 0; b <= n; b++)
37                     for(int c = 0; c <= n; c++)
38                         for(int d = 0; d <= n; d++)
39                             for(int e = 0; e <= n; e++)
40                                 dp[cur][j][a][b][c][d][e] = 0;
41 
42         for(int j = 0; j <= 5; j++) {
43             for(int a = 0; a <= n; a++) {
44                 for(int b = 0; b <= n; b++) {
45                     for(int c = 0; c <= n; c++) {
46                         for(int d = 0; d <= n; d++) {
47                             for(int e = 0; e <= n; e++) {
48                                 if(!dp[cur ^ 1][j][a][b][c][d][e]) continue;
49                                 if(e) {
50                                     LL cnt = e - (5 == j);
51                                     add(dp[cur][4][a][b][c][d + 1][e - 1], 1ll * dp[cur ^ 1][j][a][b][c][d][e] * cnt % mod);
52                                 }
53 
54                                 if(d) {
55                                     LL cnt = d - (4 == j);
56                                     add(dp[cur][3][a][b][c + 1][d - 1][e], 1ll * dp[cur ^ 1][j][a][b][c][d][e] * cnt % mod);
57                                 }
58 
59                                 if(c) {
60                                     LL cnt = c - (3 == j);
61                                     add(dp[cur][2][a][b + 1][c - 1][d][e], 1ll * dp[cur ^ 1][j][a][b][c][d][e] * cnt % mod);
62                                 }
63 
64                                 if(b) {
65                                     LL cnt = b - (2 == j);
66                                     add(dp[cur][1][a + 1][b - 1][c][d][e], 1ll * dp[cur ^ 1][j][a][b][c][d][e] * cnt % mod);
67                                 }
68 
69                                 if(a) {
70                                     LL cnt = a - (1 == j);
71                                     add(dp[cur][0][a - 1][b][c][d][e], 1ll * dp[cur ^ 1][j][a][b][c][d][e] * cnt % mod);
72                                 }
73                             }
74                         }
75                     }
76                 }
77             }
78         }
79     }
80     int ans = dp[cur][0][0][0][0][0][0];
81     printf("%d\n", ans);
82     return 0;
83 }
84 /*
85 */

 

posted @ 2018-05-29 16:47  NotNight  阅读(...)  评论(... 编辑 收藏