中石油训练赛 - The King’s Ups and Downs(记忆化搜索)

题目描述

The king has guards of all different heights. Rather than line them up in increasing or decreasing height order, he wants to line them up so each guard is either shorter than the guards next to him or taller than the guards next to him (so the heights go up and down along the line). For example, seven guards of heights 160, 162, 164, 166, 168, 170 and 172 cm. could be arranged as:

or perhaps:

The king wants to know how many guards he needs so he can have a different up and down order at each changing of the guard for rest of his reign. To be able to do this, he needs to know for a given number of guards, n, how many different up and down orders there are:
For example, if there are four guards: 1, 2, 3, 4 can be arranged as:

1324, 2143, 3142, 2314, 3412, 4231, 4132, 2413, 3241, 1423

For this problem, you will write a program that takes as input a positive integer n, the number of guards and returns the number of up and down orders for n guards of differing heights.

输入

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow Each data set consists of single line of input containing two integers. The first integer, D is the data set number. The second integer, n (1 ≤ n ≤ 20), is the number of guards of differing heights.

输出

For each data set there is one line of output. It contains the data set number (D) followed by a single space, followed by the number of up and down orders for the n guards.

样例输入

样例输出

1 1
2 4
3 10
4 740742376475050

题目链接：点击查看

题目大意：给定n个士兵，身高分别为1~n，各不相同，问符合条件的排列有多少种，符合条件的排列为：

奇数位的士兵全部高于偶数位的士兵
奇数位的士兵全部低于偶数位的士兵

满足上述两个条件的任意一条即视为符合条件

题目分析：这个题目和之前做过的一道数位dp很像，叫Mountain number，也是让求一定区间范围内满足条件的数字有多少种，那么我们可以借助这个思想，利用记忆化搜索来做，因为如果只是单纯的暴力递归，那么需要遍历n的阶乘的时间复杂度，20的阶乘就到了1e19的程度了，所以暴力枚举肯定会超时，所以我们选用记忆化搜索，举个很简单的例子，如果n为4，那么其中的两种情况为1234和1243，显而易见，前两位的12被枚举了两次，如果n变大，那么重复递归的时间复杂度呈指数级别上升，多做了很多无用的操作，所以我们采取记忆化，类似于数位dp的记忆化，开一个数组dp[pos][pre][state]，pos代表的是位数，pre代表的是前面的一个数，state代表的是该上升还是该下降，这样就能将前两位为12的次数记下来，然后在和后面两位为34和43的相加，这样需要枚举的次数大大降低，就可以再规定范围内实现这个题目了，上代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<sstream>
using namespace std;typedef long long LL;const int inf=0x3f3f3f3f;const int N=1e5+100;int n;LL dp[25][25][2];//pos,prebool vis[25];LL ans=0;LL dfs(int pos,int pre,bool state)//state:0表示该上升，1表示该下降
{if(pos==0)return 1;if(dp[pos][pre][state]!=-1)return dp[pos][pre][state];LL ans=0;for(int i=1;i<=n;i++){if(vis[i])continue;if(state&&pre>i){vis[i]=true;ans+=dfs(pos-1,i,!state);vis[i]=false;}else if(!state&&pre<i){vis[i]=true;ans+=dfs(pos-1,i,!state);vis[i]=false;}}return dp[pos][pre][state]=ans;
}int main()
{
//  freopen("input.txt","r",stdin)int w;cin>>w;while(w--){int num;memset(dp,-1,sizeof(dp));cin>>num>>n;if(n==1)//1记得要特判{printf("%d 1\n",num);continue;}memset(vis,false,sizeof(vis));cout<<num<<' '<<dfs(n,-1,0)+dfs(n,25,1)<<endl;}return 0;
}