Vacations -CodeForces-698A

最新推荐文章于 2020-05-22 17:35:37 发布

原创最新推荐文章于 2020-05-22 17:35:37 发布 · 451 阅读

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文章标签：

#动态规划 #Vacations #CodeForces-698A #ACM入门 #WHU寒假训练

本文介绍了一个关于假期活动规划的问题，通过动态规划算法找到最少的休息天数。输入为假期天数及每天的活动选项，输出为最小休息天数。

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题目：

Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:

on this day the gym is closed and the contest is not carried out;
on this day the gym is closed and the contest is carried out;
on this day the gym is open and the contest is not carried out;
on this day the gym is open and the contest is carried out.

On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).

Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive [adj. 连贯的；连续不断的]days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.

Input

The first line contains a positive [正数]integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.

The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:

ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.

Output

Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:

to do sport on any two consecutive days,
to write the contest on any two consecutive days.

Examples

input
4
1 3 2 0

output
2

input
7
1 3 3 2 1 2 3

output
0

input
2
2 2

output
1

Note

In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.

In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.

In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.

解释一下题意：

刚入门，看英文题目有困难，边读边查单词。

Vasyay有n天的假期。。。

主要看Input

第一行：输入一个正数n，代表假期天数；

第二行：输入n个正数a1..ai...an，每个数范围[0,3],

0-----只能休息

1-----比赛或者休息

2-----锻炼或者休息

3-----比赛或者锻炼或者休息

输出：

求休息时间的最小值，这里有两个限制条件，不能连续两天锻炼或者连续两天比赛；

动态规划DP

dp[i][0]表示第i天是0的最小休息天数。

dp[i][1]表示第i天是1的最小休息天数。

dp[i][2]表示第i天是2的最小休息天数。

状态转移方程：

dp[i][0]=min(min(dp[i-1][0],dp[i-1][1]),dp[i-1][2])+1;

这是综合写法，如果看不明白，可以分成三步；

dp[i][0]=min(dp[i-1][0],dp[i-1][1]);

dp[i][0]=min(dp[i][0],dp[i-1][2]);

dp[i][0]++;因为第i天是状态0，所以天数+1

dp[i][1]=min(min(dp[i-1][0],dp[i-1][2]),dp[i-1][1]+1)+0;

特意写了一个+0，这样就很容易看明白了

如果第i天是状态1，如果前一天也是状态1，那么今天就会休息，天数+1;

dp[i][2]=min(min(dp[i-1][0],dp[i-1][1]),dp[i-1][2]+1)+0;

同理；

AC代码

#include
#include
#include//包含max min 函数 
using namespace std; 

const int INF=0x3f3f3f3f;
const int MAX=105;
int n,a[MAX],dp[MAX][4];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
	}
	memset(dp,INF,sizeof(dp));
//	if(a[1]==0) 
	dp[1][0]=1;
	if(a[1]==1) dp[1][1]=0;
	if(a[1]==2)	dp[1][2]=0;
	if(a[1]==3) dp[1][1]=dp[1][2]=0;
	for(int i=2;i<=n;i++)
	{
		//if(a[i]==0) 
		/*这句if和下面那句if都是第一次写代码写上去的，结果没过
		原因是，每一天都可以选择休息或者锻炼或者竞赛，
		所以每次循环都要记录当前状态为0的时候的最小天数
		所以不需要if判定 
		*/
		dp[i][0]=min(min(dp[i-1][0],dp[i-1][1]),dp[i-1][2])+1;
		if(a[i]==1) dp[i][1]=min(min(dp[i-1][0],dp[i-1][2]),dp[i-1][1]+1)+0;
		if(a[i]==2) dp[i][2]=min(min(dp[i-1][0],dp[i-1][1]),dp[i-1][2]+1)+0;
		if(a[i]==3) 
		{
			dp[i][1]=min(min(dp[i-1][0],dp[i-1][2]),dp[i-1][1]+1)+0;
			dp[i][2]=min(min(dp[i-1][0],dp[i-1][1]),dp[i-1][2]+1)+0;
		}
	}	
	printf("%d\n",min(min(dp[n][0],dp[n][1]),dp[n][2]));
	return 0;
}

参考：http://blog.youkuaiyun.com/w446506278/article/details/51965690，博主写的很简洁，代码中出现了个别错误，本人加上了一些注释，希望能加深理解。新手入门，写给自己看。如果有问题可以一起交流啊~

枚举法也能做，这里就不写代码了。