杭電多校第10場 6887 Task Scheduler（DP）

Problem Description
For a given permutation a1,a2,⋯,an of length n, we defined the neighbor sequence b of a, the length of which is n−1, as following:

bi={01ai<ai+1ai>ai+1
.

For example, the neighbor sequence of permutation 1,2,3,6,4,5 is 0,0,0,1,0.

Now we give you an integer n and a sequence b1,b2,⋯,bn−1 of length n−1, you should calculate the number of permutations of length n whose neighbor sequence equals to b.

To avoid calculation of big number, you should output the answer module 109+7.

Input
The first line contains one positive integer T (1≤T≤50), denoting the number of test cases. For each test case:

The first line of the input contains one integer n,(2≤n≤5000).

The second line of the input contains n−1 integer: b1,b2,⋯,bn−1

There are no more than 20 cases with n>300.

Output
For each test case:

Output one integer indicating the answer module 109+7.

Sample Input
2
3
1 0
5
1 0 0 1

Sample Output
2
11

題意：
求多少種前n個數的排列符合題目中的限制

思路：
定義$f[i][j]$為考慮了前$i$個數的排列，第$i$個位置放的數在前$i$個數中為$j$的小方案數。
那麼如果$a[i]>a[i-1]，$也就是$b[i-1]=0$。
$f[i][j]=\sum f[i-1][k]，1\le k\le j-1$
$j$可以等於$k$，因為前$i$個數的第$j$小一定小於前$i-1$個數的第$j$小。下面$j$可以等於$k$同理。

如果$a[i]<a[i-1]，$也就是$b[i-1]=1$。
$f[i][j]=\sum f[i-1][k]，j\le k\le i-1$

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;
const int maxn = 5005;
const int mod = 1e9 + 7;

ll f[maxn][maxn];
int a[maxn];

int main() {
    int T;scanf("%d",&T);
    while(T--) {
        int n;scanf("%d",&n);
        for(int i = 1;i <= n;i++) {
            for(int j = 1;j <= n;j++) {
                f[i][j] = 0;
            }
        }
        for(int i = 1;i < n;i++) scanf("%d",&a[i]);
        f[1][1] = 1;
        
        for(int i = 2;i <= n;i++) {
            ll sum = 0;
            if(a[i - 1] == 0) { //a[i]>a[i-1]
                for(int j = 1;j <= i;j++) {
                    f[i][j] = sum;
                    sum = (sum + f[i - 1][j]) % mod;
                }
            } else { //a[i]<a[i-1]
                for(int j = i - 1;j >= 1;j--) {
                    sum = (sum + f[i - 1][j]) % mod;
                    f[i][j] = sum;
                }
            }
            
        }
        ll ans = 0;
        for(int i = 1;i <= n;i++) {
            ans = (ans + f[n][i]) % mod;
        }
        printf("%lld\n",ans);
    }
    return 0;
}