计蒜网 Candy （模拟&技巧）

Candy

There are $N$ children standing in a line. Each child is assigned a rating value.

You are giving candies to these children subjected to the following requirements:

(1) Each child must have at least one candy.

(2) Children with a higher rating get more candies than their neighbors.

What is the minimum candies you must give?

Input:

The input consists of multiple test cases.

The first line of each test case has a number $N$, which indicates the number of students.

Then there are $N$ students rating values, $1\le N\le 300,1\le values\le 10000$.

Output:

The minimum number of candies you must give.

样例1

输入：

51 2 3 4 55 1 3 5 3 6

输出：

159

//题意：输入n，接下来有n个数

表示n个人的身高，现在规定高的人得到的蜡烛数要比他旁边矮的人的蜡烛数多，问最少的蜡烛数是多少？

//思路：

可以从头开始判断，一直找连续的递减的序列，找到后就将这个递减序列赋值，表示他获得的蜡烛数，然后再继续向后找，直到找完为止。

最后累加求和即可。

赋值的时候有技巧，得注意，看代码

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;int a[310];int dp[310];int main(){int n,i,j,k;while(scanf("%d",&n)!=EOF){memset(dp,0,sizeof(dp));dp[0]=1;int x,y;int sum=0;for(i=0;i<n;i++)scanf("%d",&a[i]);k=a[0];int cnt=1,m=0,s=0,e,kk;for(i=1;i<n;i++){if(k>=a[i]){cnt++;if(k==a[i])m++;}else{e=i;if(cnt==1)dp[i-1]=dp[i-2]+1;else{if(cnt-m<dp[s-1]+1)sum+=dp[s-1]+1-(cnt-m);dp[s]=cnt-m;kk=a[s];for(j=s+1;j<e;j++){if(a[j]==kk)dp[j]=dp[j-1];elsedp[j]=dp[j-1]-1;kk=a[j];}}s=e;cnt=1;m=0;}k=a[i];}if(cnt==1)dp[n-1]=dp[n-2]+1;else{if(cnt-m<dp[s-1]+1)sum+=dp[s-1]+1-(cnt-m);dp[s]=cnt-m;kk=a[s];for(j=s+1;j<n;j++){if(a[j]==kk)dp[j]=dp[j-1];elsedp[j]=dp[j-1]-1;kk=a[j];}}for(i=0;i<n;i++)sum+=dp[i];printf("%d\n",sum);}return 0;}