Bzoj 1079: [SCOI2008] Coloring scheme (DP)

Dp... Or a memory search ....

Note that 1 <= c_i <= 5, although 1 <= k <= 15, but we can take the remaining paint number the same as the equivalence class

DP (A, B, C, D, E, p) indicates that the paint with paint number 1 is a, and the paint with 2 paints has a B ... p indicates which paint color was used for the last block of wood. Then move on to the code bar ...

--------------------------------------------------------------------

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <iostream>#define CLR (x, C) memset (x, C, sizeof (x))#define REP (i, n) for (int i = 0; i < n; ++i)#define MOD (x) ((x)%= MoD)using namespace std;const int MOD = 1e9 + 7;const int MAXN = n; int h[MAXN] [MAXN] [MAXN] [MAXN] [MAXN] [MAXN];int dfs (int a, int b, int c, int d, int e, int p) {int &ans = h[a [b] [C] [d] [E] [P];if (ans) return ans;ans = 0;if (a) mod (ans + = 1LL * (A-(P = = 2)) * DFS (A-1, B, C, D, e, 1)% mod);if (b) mod (ans + + 1LL * (b-(P = = 3)) * DFS (A + 1, b-1, C, D, E, 2)% mod);if (c) mod (ans + + 1LL * (c-(P = = 4)) * DFS (A, B + 1, c-1, D, E, 3)% mod);if (d) mod (ans + = 1LL * (d-(P = = 5)) * DFS (A, B, C + 1, d-1, E, 4)% mod);if (e) mod (ans + = 1LL * E * DFS (A, B, C, D + 1, e-1, 5)% mod);return ans;}int main () {freopen ("test.in", "R", stdin);CLR (h, 0);Rep (i, 6) h[0 [0] [0] [0] [0] [i] = 1;int k, t[6] = {0, 0, 0, 0, 0, 0};cin >> K;While (k--) {int v;scanf ("%d", &v);t[v]++;}cout << dfs (t[1], t[2], t[3], t[4], t[5], 0) << "\ n";return 0;} 

-------------------------------------------------------------------- 

1079: [SCOI2008] Coloring scheme time limit: ten Sec Memory Limit: 162 MB
Submit: 1100 Solved: 697
[Submit] [Status] [Discuss] Description

There are n pieces of wood lined up, numbered from left to right in 1~n. You have a k color of paint, wherein the first color of the paint is enough to tu ci a wood block. All the paint is just enough to fill all the pieces of wood, namely c1+c2+...+ck=n. Adjacent two pieces of wood painted the same color looks very ugly, so you want to count any two adjacent wood color different coloring scheme.

Input

The first behavior is a positive integer k, the second line contains k integers c1, c2, ..., CK.

Output

Outputs an integer, which is the result of modulo 1,000,000,007 for the total number of scenarios.

Sample Input3
1 2 3Sample OutputTenHINT

100% of data meet: 1 <= k <=, 1 <= ci <= 5

Source

Bzoj 1079: [SCOI2008] Coloring scheme (DP)