Pku-3250 Bad Hair Day_poj3250time limit exceeded-优快云博客

本文链接：https://blog.youkuaiyun.com/kimixuchen/article/details/8799574

解决一个算法问题，计算每头牛能看到前方多少头比自己矮的牛，通过两种不同的算法实现，一种是逐个比较，另一种是利用递减栈优化。

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3250 - Bad Hair Day

时间限制:	2000MS
内存限制:	65536K

问题描述

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 ofc₁ through c_N.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

输入说明

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.

输出说明

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

输入样例

输出样例

来源

N/A

//首先奉上一种解法，本人制作，之后还有一种大神解法

#include <iostream>
using namespace std;

const int N=80005;
int h[N];
int num[N],maxx[N];		//num[i],第i个牛可以看见的头数目。 maxx[i],在第i头牛右边第一头比i高的牛序号

int main(){
    int n,k;
    unsigned long long ans = 0;	//注意是 long long
    while(cin>>n){
        ans=0;
        for(int i=0; i<n; i++){
            cin>>h[i];

        }
        num[n-1]=0; maxx[n-1]=-1;

        for(int i=n-2; i>=0; i--){
            if(h[i]<h[i+1]){
                num[i]=0; maxx[i]=i+1;          
            }
            else{
                k=i+1;
                while(k!=-1 && h[k]<h[i]){
                    num[i]+=(num[k]+1);
                    k=maxx[k];
                }
                maxx[i]=k;
                ans+=num[i];
            }



        }
    cout<<ans<<endl;
    }

return 0;
}

//接下来是完爆上面解法的大神解法

#include <stdio.h>


int a[80010], n, top;
int main() {
scanf("%d", &n);
long long ans = 0;
while (n --) {
int x; scanf("%d", &x);
while (top && a[top-1] <= x) top --;	//a[]为递减数列，而且a[]中元素个数即为比当前牛高度高的左边牛数目
ans += top;
a[top++] = x;
}
printf("%lld\n", ans);
return 0;