BZOJ 1996 合唱队(DP)

考虑从最后的队形开始依次还原最初的队形。

对于当前的队形，要么选最左边的，要么选最右边的。 如果选了左边的，那么下次选择的一定是大于它的。右边的同理。

所以定义dp[mark][l][r]为区间[l,r]的选择状态为mark的方法数。

然后记忆化搜索一下就可以了。

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 19650827
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=1005;
//Code begin...

int dp[2][N][N], a[N];

int dfs(int mark, int l, int r){
    if (~dp[mark][l][r]) return dp[mark][l][r];
    if (l==r) return dp[mark][l][r]=1;
    int res=0;
    if (mark) {
        if (a[l]<a[r]) res+=dfs(0,l,r-1);
        if (r-1!=l&&a[r-1]<a[r]) res+=dfs(1,l,r-1);
    }
    else {
        if (a[r]>a[l]) res+=dfs(1,l+1,r);
        if (l+1!=r&&a[l+1]>a[l]) res+=dfs(0,l+1,r);
    }
    return dp[mark][l][r]=res%MOD;
}
int main ()
{
    mem(dp,-1);
    int n;
    scanf("%d",&n);
    FOR(i,1,n) scanf("%d",a+i);
    printf("%d
",(dfs(0,1,n)+dfs(1,1,n))%MOD);
    return 0;
}