HDOJ 4689 Derangement DP

DP详解见:

http://blog.csdn.net/liguan1/article/details/10468139

Derangement

Time Limit: 7000/7000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 846    Accepted Submission(s): 256

Problem Description

A derangement is a permutation such that none of the elements appear in their original position. For example, [5, 4, 1, 2, 3] is a derangement of [1, 2, 3, 4, 5]. Subtracting the original permutation from the derangement, we get the derangement difference [4, 2, -2, -2, -2], where none of its elements is zero. Taking the signs of these differences, we get the derangement sign [+, +, -, -, -]. Now given a derangement sign, how many derangements are there satisfying the given derangement sign?

 

Input

There are multiple test cases. Process to the End of File.
Each test case is a line of derangements sign whose length is between 1 and 20, inclusively.

 

Output

For each test case, output the number of derangements.

 

Sample Input

+-++---

 

Sample Output

113

 

Author

Zejun Wu (watashi)

 

Source

2013 Multi-University Training Contest 9

 

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;typedef long long int LL;LL dp[50][50];char str[50];int main(){    while(cin>>str)    {        if(str[0]=='-')        {            puts("0"); continue;        }        memset(dp,0,sizeof(dp));        int n=strlen(str);        dp[1][1]=1;        for(int i=2;i<=n;i++)        {            for(int j=0;j<=i;j++)            {                if(str[i-1]=='+')                {                    if(j) dp[i][j]+=dp[i-1][j-1];                    dp[i][j]+=dp[i-1][j]*j;                }                else if(str[i-1]=='-')                {                    dp[i][j]+=dp[i-1][j]*j;                    dp[i][j]+=dp[i-1][j+1]*(j+1)*(j+1);                }            }        }        cout<<dp[n][0]<<endl;    }    return 0;}