HDU4055 Number String（计数dp）

Number String

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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 length n that contains each of the integers 1 through n 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.

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题意

给定一个1-n的排列，$I$表示比前一个大，$D$表示比前一个小，？可大可小，问有多少种排列满足条件。

分析

定义$dp[\,i\,][\,j\,]$表示前i位序列末尾是j的序列共有多少个。
输入时用scanf("%s",A+2);会更方便。
那么，当这个字符比前一个大时(A[i]=='I'):

$$dp[\,i\,][\,j\,]\!=\!dp[\,i\,][\,j\!-\!1\,]\!+\!dp[\,i\!-\!1\,][\,j\!-\!1\,]$$
当 A[i]=='D'时:
$$dp[\,i\,][\,j\,]\!=\!dp[\,i\,][\,j\!+\!1\,]\!+\!dp[\,i\!-\!1\,][\,j\!+\!1\,]$$
当 A[i]=='?'时:
$$dp[\,i\,][\,j\,]\!=\!\sum_{x=1}^{i-1}dp[\,i\!-\!1\,][\,x\,]$$
显然边界为 $dp[\,1\,][\,1\,]\!=\!1.$
时间复杂度 $O(n^2T)$（ $T$为数据组数）

参考

CODE

#include<cstdio>
#include<cstring>
#include<memory.h>
#define mod 1000000007
#define N 1005
#define FOR(i,a,b) for(int i=(a),i##_END_=(b);i<=i##_END_;i++)
#define ROF(i,a,b) for(int i=(a),i##_END_=(b);i>=i##_END_;i--)
char a[N];
int dp[N][N];
//dp[i][j]表示处理完第i位，序列末尾j的序列共有多少个。
inline int Add(int x,int y) {
    int t=x+y;
    if(t>=mod)t-=mod;
    return t;
}

int main() {
    while(~scanf("%s",a+2)) {
        int n=strlen(a+2)+1;
        memset(dp,0,sizeof(dp));
        dp[1][1]=1;
        FOR(i,2,n) {
            if(a[i]=='D')ROF(j,i-1,1)
                dp[i][j]=Add(dp[i][j+1],dp[i-1][j]);
            else if(a[i]=='I')FOR(j,2,i)
                dp[i][j]=Add(dp[i][j-1],dp[i-1][j-1]);
            else {
                int sum=0;
                FOR(j,1,i-1)sum=Add(sum,dp[i-1][j]);
                FOR(j,1,i)dp[i][j]=sum;
            }
        }
        int ans=0;
        FOR(i,1,n)ans=Add(ans,dp[n][i]);
        printf("%d\n",ans);
    }
    return 0;
}

当然因为每层dp都是从上一层转移所以可以优化空间。
有这个时间干嘛不多写几题

#include<cstdio>
#include<cstring>
#include<memory.h>
#define mod 1000000007
#define N 1005
#define FOR(i,a,b) for(int i=(a),i##_END_=(b);i<=i##_END_;i++)
#define ROF(i,a,b) for(int i=(a),i##_END_=(b);i>=i##_END_;i--)
char a[N];
int dp[2][N];
inline int Add(int x,int y) {
    int t=x+y;
    if(t>=mod)t-=mod;
    return t;
}

int main() {
    while(~scanf("%s",a+2)) {
        int n=strlen(a+2)+1;
        memset(dp,0,sizeof(dp));
        dp[1][1]=1;
        FOR(i,2,n) {
            int cur=i&1;
            if(a[i]=='D') {
                dp[cur][i]=0;
                ROF(j,i-1,1)
                    dp[cur][j]=Add(dp[cur][j+1],dp[!cur][j]);
            } else if(a[i]=='I') {
                dp[cur][1]=0;
                FOR(j,2,i)
                    dp[cur][j]=Add(dp[cur][j-1],dp[!cur][j-1]);
            } else {
                int sum=0;
                FOR(j,1,i-1)sum=Add(sum,dp[!cur][j]);
                FOR(j,1,i)dp[cur][j]=sum;
            }
        }
        int ans=0;
        FOR(i,1,n)ans=Add(ans,dp[n&1][i]);
        printf("%d\n",ans);
    }
    return 0;
}