# BZOJ 1079: [SCOI2008]着色方案 记忆化搜索

# 1079: [SCOI2008]着色方案

Time Limit: 20 Sec

Memory Limit: 256 MB

## 题目连接

http://www.lydsy.com/JudgeOnline/problem.php?id=1079

## Description

Under two situations the player could score one point.

⋅1.

**If you touch a buoy before your opponent, you will get one point.**For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won't get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.

⋅2.

**Ignoring the buoys and relying on dogfighting to get point.**If you and your opponent meet in the same position, you can try to fight with your opponent to score one point. For the proposal of game balance, two players are not allowed to fight before buoy #2 is touched by anybody.

There are three types of players.

**Speeder**: As a player specializing in high speed movement, he/she tries to avoid dogfighting while attempting to gain points by touching buoys.

**Fighter**: As a player specializing in dogfighting, he/she always tries to fight with the opponent to score points. Since a fighter is slower than a speeder, it's difficult for him/her to score points by touching buoys when the opponent is a speeder.

**All-Rounder**: A balanced player between Fighter and Speeder.

There will be a training match between Asuka (All-Rounder) and Shion (Speeder).

Since the match is only a training match, the rules are simplified:

**the game will end after the buoy**#1

**is touched by anybody**. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.

Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and

**the opponent will be stunned for T seconds after dogfighting.**Since Asuka is slower than Shion, she decides to fight with Shion for only one time during the match. It is also assumed that if Asuka and Shion touch the buoy in the same time, the point will be given to Asuka and Asuka could also fight with Shion at the buoy. We assume that in such scenario, the dogfighting must happen after the buoy is touched by Asuka or Shion.

The speed of Asuka is V

**Input**

第一行为一个正整数k，第二行包含k个整数c1, c2, ... , ck。

**Output**

输出一个整数，即方案总数模1,000,000,007的结果。

**Sample Input**

`3`

1 2 3

**Sample Output**

`10`

## HINT

**题意**

**题解：**

直接dp[a][b][c][d][e][f]表示上一个状态为f时，我可以涂1次的有a个，涂两次的有b个，涂三次的有c个，涂四次的有d个，涂五次的有e个的方案数

直接记忆化搜索转移就好了

**代码**

#include<iostream> #include<stdio.h> using namespace std; int tmp[10]; int dp[16][16][16][16][16][7]; int vis[16][16][16][16][16][7]; #define mod 1000000007 long long dfs(int a,int b,int c,int d,int e,int n){ if(vis[a][b][c][d][e][n]) return dp[a][b][c][d][e][n]; if(a+b+c+d+e==0) return dp[a][b][c][d][e][n]=1; long long ans=0; if(a)ans+=(a-(n==2))*dfs(a-1,b,c,d,e,1); if(b)ans+=(b-(n==3))*dfs(a+1,b-1,c,d,e,2); if(c)ans+=(c-(n==4))*dfs(a,b+1,c-1,d,e,3); if(d)ans+=(d-(n==5))*dfs(a,b,c+1,d-1,e,4); if(e)ans+=(e-(n==6))*dfs(a,b,c,d+1,e-1,5); vis[a][b][c][d][e][n]=1; ans%=mod; return dp[a][b][c][d][e][n]=ans; } int main() { int n; scanf("%d",&n); for(int i=0;i<n;i++) { int x;scanf("%d",&x); tmp[x]++; } printf("%d\n",dfs(tmp[1],tmp[2],tmp[3],tmp[4],tmp[5],0)); }