BZOJ1996 [Hnoi2010]chorus 合唱队

很容易想到区间DP

然后发现这个区间只和圆序列的最后一个数有关，而原序列的最后一个数只可能是现在区间的头或者尾

令$f[i][j][0/1]$表示在区间$[i, j]$之间，原序列的最后一个数是当前区间的头/尾的总答案数

于是只要讨论$a[i], a[i + 1], a[j - 1], a[j]$之间的关系搞一搞就可以了

 /**************************************************************
     Problem: 1996
     User: rausen
     Language: C++
     Result: Accepted
     Time:112 ms
     Memory:8700 kb
 ****************************************************************/

 #include <cstdio>
 #include <cstring>

 using namespace std;
 const int N = (int) 1e3 + ;
 const int mod = ;

 int n, a[N];
 int f[N][N][], ans;

 int main() {
     int i, j, len;
     scanf("%d", &n);
     if (n == ) {
         puts("");
         return ;
     }
     for (i = ; i <= n; ++i) scanf("%d", a + i);
     for (i = ; i < n; ++i)
         f[i][i + ][] = f[i][i + ][] = bool(a[i] < a[i + ]);
     for (len = ; len <= n; ++len)
         for (i = ; i <= n - len + ; ++i) {
             j = i + len - , f[i][j][] = f[i][j][] = ;
             if (a[j] > a[j - ]) f[i][j][] += f[i][j - ][];
             if (a[j] > a[i]) f[i][j][] += f[i][j - ][];
             if (a[i] < a[i + ]) f[i][j][] += f[i + ][j][];
             if (a[i] < a[j]) f[i][j][] += f[i + ][j][];
             f[i][j][] %= mod, f[i][j][] %= mod;
         }
     ans = (f[][n][] + f[][n][]) % mod;
     printf("%d\n", ans);
     return ;
 }