【CF - 699C】 Vacations (日程安排 dp)
題干:
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 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.
Examples
Input
4 1 3 2 0Output
2Input
7 1 3 3 2 1 2 3Output
0Input
2 2 2Output
1Note
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.
?
題目大意:
? ? Vasya在n天中,有三件事情可以做,健身、寫作或者休息,但是健身和寫作不能連續兩天都去做,但是連續休息兩天是允許的,問題是在這n天中,Vasya最少可以休息幾天?
題意2::Vasyayo有n天的假期,他每天可以選擇運動,比賽或者休息。但每天都四個狀態,體育場開門沒有比賽,體育場關門有比賽,體育場開門有比賽,體育場關門沒比賽。問他最少休息多少天?
解題報告:
? ? ?因為題目中顯然有狀態的轉移,所以考慮dp
AC代碼:
#include<bits/stdc++.h>using namespace std; const int INF = 0x3f3f3f3f; int dp[1000][4];//1是休息,2是比賽,3是健身 第二位代表當天做了什么(所以首先得能選擇做這個決策,所以m=1時不需要更新 健身那一層),而不是 int a[1000]; int min(int x,int y,int z) {return min(x,min(y,z)); } int main() {int n;int minn ;cin>>n;for(int i = 1; i<=n; i++) {cin>>a[i];//1是比賽,2是健身 }//看我當天能夠做什么事來選擇更新什么值 for(int i = 1; i<=n; i++) {if(a[i] == 0) {minn = INF;for(int k = 1; k<=3; k++) minn = min(minn,dp[i-1][k]); // printf("minn = %d\n",minn);dp[i][1] = minn+1;dp[i][2] = dp[i][3] = INF;}else if(a[i] == 1) {dp[i][2] = min(dp[i-1][1],dp[i-1][3]);dp[i][1] = min(dp[i-1][1] + 1,dp[i-1][2]+1,dp[i-1][3] + 1);dp[i][3] = INF;}else if(a[i] == 2) {dp[i][1] = min(dp[i-1][1]+1,dp[i-1][2]+1,dp[i-1][3]+1);dp[i][2] = INF;dp[i][3] = min(dp[i-1][2],dp[i-1][1]);}else {dp[i][1] = min(dp[i-1][1]+1,dp[i-1][2]+1,dp[i-1][3]+1);dp[i][2] = min(dp[i-1][1],dp[i-1][3]);dp[i][3] = min(dp[i-1][1],dp[i-1][2]);}} // for(int i = 1; i<=n; i++)printf("%d\n",min(dp[n][1],dp[n][2],dp[n][3]));return 0 ;}總結
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