洛谷 P2947 [USACO09MAR]向右看齐Look Up (队列)
题目描述
Farmer John’s N (1 <= N <= 100,000) cows, conveniently numbered 1..N, are once again standing in a row. Cow i has height H_i (1 <= H_i <= 1,000,000).
Each cow is looking to her left toward those with higher index numbers. We say that cow i ‘looks up’ to cow j if i < j and H_i < H_j. For each cow i, FJ would like to know the index of the first cow in line looked up to by cow i.
Note: about 50% of the test data will have N <= 1,000.
约翰的N(1≤N≤10^5)头奶牛站成一排,奶牛i的身高是Hi(l≤Hi≤1,000,000).现在,每只奶牛都在向右看齐.对于奶牛i,如果奶牛j满足i < j且Hi < Hj,我们可以说奶牛i可以仰望奶牛j. 求出每只奶牛离她最近的仰望对象.
Input
输入输出格式
输入格式:
* Line 1: A single integer: N
- Lines 2..N+1: Line i+1 contains the single integer: H_i
第 1 行输入 N,之后每行输入一个身高 H_i。
输出格式:
* Lines 1..N: Line i contains a single integer representing the smallest index of a cow up to which cow i looks. If no such cow exists, print 0.
共 N 行,按顺序每行输出一只奶牛的最近仰望对象,如果没有仰望对象,输出 0。
输入输出样例
输入样例#1:
6
3
2
6
1
1
2
输出样例#1:
3
3
0
6
6
0
说明
FJ has six cows of heights 3, 2, 6, 1, 1, and 2.
Cows 1 and 2 both look up to cow 3; cows 4 and 5 both look up to cow 6; and cows 3 and 6 do not look up to any cow.
【输入说明】6 头奶牛的身高分别为 3, 2, 6, 1, 1, 2.
【输出说明】奶牛#1,#2 仰望奶牛#3,奶牛#4,#5 仰望奶牛#6,奶牛#3 和#6 没有仰望对象。
【数据规模】
对于 20%的数据: 1≤N≤10;
对于 50%的数据: 1≤N≤1,000;
对于 100%的数据:1≤N≤100,000;1≤H_i≤1,000,000;
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分析
单调队列模板打法,因为head为1,所以要保证tail>0,当目前扫到的数大于队列末尾的数时,这个数一定是队列末尾的数,右端第一个比它大的数,所以就把这个这个数储存入队列末尾数的an数组中,并将tail–,即把末尾数移除队列,紧接着比较队列中下个数,直到遇到第一个≥这个数或队列空为止。 总的核心思想就是维护一个不上升连续序列,在维护的同时找到答案。
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程序:
#include<iostream>
using namespace std;
int n,a[100005],q[100005],an[100005],tail=0;
int main()
{
cin>>n;
for (int i=1;i<=n;i++) cin>>a[i];
for (int i=1;i<=n;i++)
{
while (tail&&a[i]>a[q[tail]])
{
an[q[tail]]=i;
tail--;
}
q[++tail]=i;
}
for (int i=1;i<=n;i++) cout<<an[i]<<endl;
return 0;
}