ZOJ Problem Set - 3543 Number String DP

Number String
Time Limit: 5 Seconds     
Memory Limit: 65536 KB
The signature of a permutation is a string that is computed as follows: for each pair of consecutive elements of the permutation, write down the letter 'I' (increasing) if the second element is greater than the first one, otherwise write down the letter 'D' (decreasing). For example, the signature of the permutation {3,1,2,7,4,6,5} is "DIIDID".
Your task is as follows: You are given a string describing the signature of many possible permutations, find out how many permutations satisfy this signature.
Note: For any positive integer n, a permutation of n elements is a sequence of lengthn that contains each of the integers 1 throughn exactly once.

Input

Each test case consists of a string of 1 to 1000 characters long, containing only the letters 'I', 'D' or '?', representing a permutation signature.
Each test case occupies exactly one single line, without leading or trailing spaces.
Proceed to the end of file. The '?' in these strings can be either 'I' or 'D'.

Output

For each test case, print the number of permutations satisfying the signature on a single line. In case the result is too large, print the remainder modulo 1000000007.

Sample Input

II
ID
DI
DD
?D
??

Sample Output

1
2
2
1
3
6

Hint

Permutation {1,2,3} has signature "II". Permutations {1,3,2} and {2,3,1} have signature "ID". Permutations {3,1,2} and {2,1,3} have signature "DI". Permutation {3,2,1} has signature "DD". "?D"can be either "ID"or "DD". "??"gives all possible permutations of length 3.
하나의 길이 n의 배열(1-n 수)은 I가 앞의 수보다 크다는 것을 의미하고, D는 뒤의 수가 앞의 수보다 작다는 것을 의미하는가?뜻대로몇 가지 배열 방법이 있습니까?
dp[i][j]는 길이가 i인 배열의 마지막 수가 j이고 전 i-1개의 관계를 만족시키는 것이 몇 가지인지를 나타낸다.현재 i위로 순환하고 있습니다. 이 분은 j를 뽑습니다. 만약 관계가 I라면 dp[i][j]=sum(dp[i-1][k], 1<=kj입니다.하면, 만약, 만약...그럼 모든 상황이 만족됩니다. dp[i][j]=sum(dp[i-1][k],1<=k<=i-1).sum를 저장하면 한 층의 순환, 복잡도 O(N^2)를 줄일 수 있습니다.

#include<cstring>
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>
#define INF 0x3f3f3f3f
using namespace std;

typedef long long LL;

const int MAXN=1010;
const LL MOD= 1000000007;

LL dp[MAXN][MAXN],sum[MAXN][MAXN];
char str[MAXN];

int main(){
    freopen("in.txt","r",stdin);
    while(scanf("%s",str+1)!=EOF){
        int len=strlen(str+1);
        memset(dp,0,sizeof(dp));
        memset(sum,0,sizeof(sum));
        dp[1][1]=1;
        sum[1][1]=1;
        for(int i=1;i<=len;i++)
            for(int j=1;j<=i+1;j++){
                if(str[i]=='I') dp[i+1][j]=sum[i][j-1];
                else if(str[i]=='D') dp[i+1][j]=(sum[i][i]-sum[i][j-1]+MOD)%MOD;
                else dp[i+1][j]=sum[i][i];
                sum[i+1][j]=(sum[i+1][j-1]+dp[i+1][j])%MOD;
            }
        printf("%lld
",sum[len+1][len+1]);
    }
    return 0;
}