1079: [SCOI2008]著色方案

Time Limit: 20 Sec

Memory Limit: 256 MB

題目連接

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

Description

有n個木塊排成一行，從左到右依次編號為1~n。你有k種顏色的油漆，其中第i種顏色的油漆足夠涂ci個木塊。所有油漆剛好足夠涂滿所有木塊，即c1+c2+...+ck=n。相鄰兩個木塊涂相同色顯得很難看，所以你希望統計任意兩個相鄰木塊顏色不同的著色方案。
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 V1?m/s. The speed of Shion is V2?m/s. Is there any possibility for Asuka to win the match (to have higher score)?

Input

第一行為一個正整數k，第二行包含k個整數c1, c2, ... , ck。

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Output

輸出一個整數，即方案總數模1,000,000,007的結果。

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Sample Input

3
1 2 3

Sample Output

10

HINT

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題意

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題解：

直接dp[a][b][c][d][e][f]表示上一個狀態為f時，我可以涂1次的有a個，涂兩次的有b個，涂三次的有c個，涂四次的有d個，涂五次的有e個的方案數

直接記憶化搜索轉移就好了

代碼

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#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)); }

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