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 + 5;
const int mod = 19650827;
  
int n, a[N];
int f[N][N][2], ans;
  
int main() {
    int i, j, len;
    scanf("%d", &n);
    if (n == 1) {
        puts("1");
        return 0;
    }
    for (i = 1; i <= n; ++i) scanf("%d", a + i);
    for (i = 1; i < n; ++i)
        f[i][i + 1][0] = f[i][i + 1][1] = bool(a[i] < a[i + 1]);
    for (len = 3; len <= n; ++len)
        for (i = 1; i <= n - len + 1; ++i) {
            j = i + len - 1, f[i][j][0] = f[i][j][1] = 0;
            if (a[j] > a[j - 1]) f[i][j][1] += f[i][j - 1][1];
            if (a[j] > a[i]) f[i][j][1] += f[i][j - 1][0];
            if (a[i] < a[i + 1]) f[i][j][0] += f[i + 1][j][0];
            if (a[i] < a[j]) f[i][j][0] += f[i + 1][j][1];
            f[i][j][0] %= mod, f[i][j][1] %= mod;
        }
    ans = (f[1][n][0] + f[1][n][1]) % mod;
    printf("%d
", ans);
    return 0;
}