Fantasy Cricket UVA

动态规划的一道题目，逆向思维，对于得到的串，‘U’代表需要向下放，‘D’代表需要向上放，dp[i][j]也就代表着当前处理到第i个字符，还有j个‘U’字符没有放下，如果当前处理的字符是‘U’，那么必须将该字符拿走，如果处理的是‘D’，必须将该字符向前放置，然后再考虑对应操作结束之后的状态转移方程即可，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;const int maxn = 1005;char s[maxn];long long dp[maxn][maxn];const long long mod = 1000000007;int main(){int T;cin >> T;for (int i = 1; i <= T; i++){scanf("%s",s+1);int length = strlen(s+1);memset(dp,0,sizeof(dp));dp[0][0] = 1;for (int i = 1; i <= length; i++){if (s[i] == 'D'){for (int j = 0; j <= i; j++){dp[i][j] = (dp[i][j] + dp[i - 1][j] * j%mod +dp[i - 1][j + 1] * (j + 1) % mod*(j + 1) % mod) % mod;}}else if (s[i] == 'U'){for (int j = 0; j <= i; j++){int a = 0;if (j > 0) a = dp[i - 1][j - 1];dp[i][j] = (dp[i][j] + a + dp[i - 1][j]*j%mod) % mod;}}else{for (int j = 0; j <= i; j++)dp[i][j] = dp[i - 1][j];}}cout <<"Case "<<i<<": "<< dp[length][0] << endl;}return 0;}