BZOJ 1996: [Hnoi2010]chorus 合唱队(dp)

简单的dp题..不能更水了..

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#define rep(i,n) for(int i=0;i<n;++i)

#define clr(x,c) memset(x,c,sizeof(x))

using namespace std;

const int maxn=1000+5;

const int mod=19650827;

int n;

int goal[maxn];

int d[maxn][maxn][2];//0放入左,1放入右

int dp(int l,int r,int op) {

int &ans=d[l][r][op];

if(ans>=0) return ans;

ans=0;

if(op) {

if(goal[l]<goal[r]) ans=dp(l,r-1,0);

if(l!=r-1 && goal[r-1]<goal[r]) (ans+=dp(l,r-1,1))%=mod;

} else {

if(goal[l+1]>goal[l]) ans=dp(l+1,r,0);

if(l+1!=r && goal[r]>goal[l]) (ans+=dp(l+1,r,1))%=mod;

}

return ans%=mod;

}

int main()

{

// freopen("test.in","r",stdin);

// freopen("test.out","w",stdout);

clr(d,-1);

scanf("%d",&n);

rep(i,n) d[i][i][0]=d[i][i][1]=1;

rep(i,n) scanf("%d",&goal[i]);

printf("%d ",(dp(0,n-1,0)+dp(0,n-1,1))%mod);

return 0;

}

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1996: [Hnoi2010]chorus 合唱队

Time Limit: 4 Sec Memory Limit: 64 MB
Submit: 1045 Solved: 672
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Description

Input

Output

Sample Input

4
1701 1702 1703 1704