URAL 1225(Flags)简单递推

最新推荐文章于 2018-05-07 09:54:30 发布

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本文讨论了一种在旗帜装饰中满足特定条件排列组合的问题，详细介绍了动态规划算法的应用及求解过程。

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1225. Flags

Time limit: 1.0 second Memory limit: 64 MB

On the Day of the Flag of Russia a shop-owner decided to decorate the show-window of his shop with textile stripes of white, blue and red colors. He wants to satisfy the following conditions:

1.Stripes of the same color cannot be placed next to each other.

2.A blue stripe must always be placed between a white and a red or between a red and a white one.

Determine the number of the ways to fulfill his wish.

Example.ForN= 3 result is following:

Input

N, the number of the stripes, 1 ≤N≤ 45.

Output

M, the number of the ways to decorate the shop-window.

Sample

input	output
3	4

ProbPlem Source:2002-2003 ACM Central Region of Russia Quarterfinal Programming Contest, Rybinsk, October 2002

题意：给出红、白、蓝三种颜色；进行涂色；

要求一：相邻的颜色不能相同。要求二：蓝色的两侧颜色不同；

思路：另dp[i]表示当旗帜数为i时合法的方法数，因为当后面加一个红色或者白色气球时，得到的总方法数为dp[i-1]，当后面加入蓝色、白色或者蓝色、红色时，总方法数为dp[i-2].

代码如下：

/**********************
* author:crazy_石头
* Pro:URAl 1225
* algorithm:dp
* Time:0ms
* Judge Status:Accepted
***********************/
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
#include <set>
#include <vector>
#include <cmath>

using namespace std;

#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define INF 1<<29

const int maxn=1000+5;
long long a[maxn],n;

int main()
{
    a[1]=a[2]=2;
    rep(i,3,45)
    a[i]=a[i-1]+a[i-2];
    while(cin>>n)
        cout<<a[n]<<endl;
    return 0;
}