HDU 4689 Derangement【DP递推】【好题】【思维题】

Derangement

Time Limit: 7000/7000 MS (Java/Others)    Memory Limit: 65535/102400 K (Java/Others)
Total Submission(s): 1170    Accepted Submission(s): 396

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

1 13

 

Author

Zejun Wu (watashi)

 

Source

2013 Multi-University Training Contest 9

题意：1到n的n个数，打乱后减去原来的数的正负号，形成一串正负号的字符串，如[1,2,3,4,5]打乱后变成[5,4,1,2,3]就是[+,+,-,-,-]。给定这样一串正负号形成的字符串，有几种排列方式。

dp[i][j]表示前i个位置中有j个空位（+号）没有确定，负号全部确定。

“+'时

要么使空位增加，要么不变

增加时，直接空位增加：dp[i][i]+=dp[i-1][j-1]

不变时，表示把当前的数填到前面未确定的j个空位中去：dp[i][j]+=dp[i-1][j]*j

‘-'时，因为必须确定

要么减少，要么不变

不变时，表示把当前数填到前面的j个空位中：dp[i][j]+=dp[i-1][j]*j

减少时，表示把当前数填到前j-1位中去，再从前j-1位中拿一个数填到当前位：dp[i][j]+=dp[i-1][j-1]*(j-1)*(j-1)

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define ll long long

ll dp[30][30];
char num[100];

int main()
{
	while (~scanf("%s", num))
	{
		int len = strlen(num);
		memset(dp, 0, sizeof(dp));
		dp[0][0] = 1;
		for (int i = 1; i <= len; i++)
		{
			if (num[i - 1] == '+')
			{
				for (int j = 0; j <= i; j++)
				{
					dp[i][j] += dp[i - 1][j - 1];
					dp[i][j] += dp[i - 1][j] * (ll)j;
				}
			}
			else
			{
				for (int j = 0; j <= i; j++)
				{
					// dp[i][j - 1] += dp[i - 1][j] * j * j;
					dp[i][j] += dp[i - 1][j] * (ll)j;
					dp[i][j] += dp[i - 1][j + 1] * (ll)(j + 1) * (ll)(j + 1);
				}
			}
		}
		// printf("%lld\n", dp[len][0]);
		cout << dp[len][0] << endl;
	}
	return 0;
}