POJ 3250 Bad Hair Day 简单DP 好题

Description

Some of Farmer John's N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows' heads.

Each cow i has a specified height h_i (1 ≤ h_i ≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.

Consider this example:

        =
=       =
=   -   =         Cows facing right -->
=   =   =
= - = = =
= = = = = =
1 2 3 4 5 6 

Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow's hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow's hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!

Let c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_N.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

Input

Line 1: The number of cows, N. 
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.

Output

Line 1: A single integer that is the sum of c₁ through c_N.

Sample Input

6
10
3
7
4
12
2

Sample Output

5





题意： 有n头牛从左到右排成一行，从左到右编号为1~n，并且牛的头是朝向右的，即每头牛只可以看到他们右边的牛。
现在给出n头牛的身高，设c[i]表示第i头牛可以看到多少头牛的头顶。
求所有c[i]的和。

简单DP


设dp[i]表示第i头牛在看的时候是被第dp[i]头牛挡住视线的，即i+1~dp[i]-1这个区间的牛都可以被第i头牛看到。
所以第i头牛看到的数目为：dp[i]-i-1




这道题也可以用栈来做。

#include<cstdio>
#include<cstring>
#include<algorithm>

const int maxn=80000+5;
const long long inf=0x3f3f3f3f;

#define LL long long

LL h[maxn];
LL dp[maxn];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&h[i]);
        }
        n++;
        h[0]=h[n]=inf;

        for(int i=n-1;i>0;i--)
        {
            int tmp=i+1;
            dp[i]=i+1;
            while(h[i]>h[tmp])
            {
                dp[i]=dp[tmp];
                tmp=dp[tmp];
            }
        }
        LL ans=0;
        for(int i=1;i<n;i++)
        {
            ans+=(dp[i]-i-1);
        }
        printf("%lld
",ans);
    }
    return 0;
}