codeforces 698A Vacations 基础dp

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 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 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.
Example
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.
Vasya有n天可以安排，0表示一定要休息，1表示可以去比赛或者休息，2表示可以去健身或者休息，3表示可以去比赛或者健身或者休息，且不能连续的两天进行竞赛或者健身，问他最少需要休息几天。

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。
转移方程：
dp[i][0]=min(min(dp[i-1][0],dp[i-1][1]),dp[i-1][2])+1;

一开始全初始化为最大

#include <bits/stdc++.h>using namespace std;int a[101010];int dp[101010][4];const int INF = 0x3f3f3f3f;int main(){    int n;    cin>>n;    memset(dp,INF,sizeof(dp));    for(int i=1;i<=n;i++)    {        scanf("%d",&a[i]);    }    dp[0][0]=0;        for(int i=1;i<=n;i++)        {                dp[i][0]=min(dp[i-1][1],dp[i-1][2]);                dp[i][0]=min(dp[i][0],dp[i-1][0])+1;            if(a[i]==1)            {                dp[i][1]=min(dp[i-1][2],dp[i-1][0]);            }            if(a[i]==2)            {                dp[i][2]=min(dp[i-1][0],dp[i-1][1]);            }            if(a[i]==3)            {                dp[i][2]=min(dp[i-1][1],dp[i-1][0]);                dp[i][1]=min(dp[i-1][2],dp[i-1][0]);            }        }        int res=1e9+7;        res=min(dp[n][0],dp[n][1]);        res=min(res,dp[n][2]);        printf("%d\n",res );}