URAL——DFS找规律——Nudnik Photographer

Description

If two people were born one after another with one second difference and one of them is a child, then the other one is a child too. We get by induction that all the people are children.

Everyone knows that the mathematical department of the Ural State University is a big family of N persons, 1, 2, 3, …, N years old respectively.

Once the dean of the department ordered a photo if his big family. There were to be present all the students of the department arranged in one row. At first the dean wanted to arrange them by their age starting from the youngest student, but than he decided that it would look unnatural. Than he advised to arrange the students as follows:

1. The 1 year old student is to sit at the left end of the row.
2. The difference in ages of every two neighbors mustn’t exceed 2 years.

The dean decided that thereby the students would seem look as they were arranged by their ages (one can hardly see the difference in ages of 25 and 27 years old people). There exist several arrangements satisfying to the requirements. Photographer didn’t want to thwart dean’s desire and made the photos of all the possible mathematical department students’ arrangements.

Input

There is the integer number N, 1 ≤ N ≤ 55.

Output

the number of photos made by the photographer.

Sample Input

4

4

Notes

If N = 4 then there are following possible arrangements: (1,2,3,4), (1,2,4,3), (1,3,2,4) and (1,3,4,2).

大意：让你排座位，第一个人必须是年龄为1的，其他必须相邻的年龄间隔不超过2，可以DFS打表过。。

说是道DP。。得到dp[n] = dp[n-1] + dp[n-3] + 1看传送门

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int vis[60];
int n;
int ans;
void dfs(int m,int num)
{
    if(num == n){
        ans++;
        return ;
    }
    for(int i = m - 2; i <= m + 2; i++){
        if(i <= 1 || i > n )
            continue;
        if(!vis[i]){
            vis[i] = 1;
            dfs(i,num+1);
            vis[i] = 0;
        }
    }
}
int main()
{
    while(~scanf("%d",&n)){
        ans = 0;
        memset(vis,0,sizeof(vis));
        vis[1] = 1;
        dfs(1,1);
        printf("%d\n",ans);
    }
    return 0;
}

AC代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[60];
int main()
{
    int n;
    dp[1] = 1;
    dp[2] = 1;
    dp[3] = 2;
    while(~scanf("%d",&n)){
        for(int i = 3; i <= n ;i++)
            dp[i] = dp[i-1] + dp[i-3] + 1;
       printf("%d\n",dp[n]);
    }
    return 0;
}