FZU 1845 Look Up(单调栈）

Description

N (1 <= N <= 100,000) monkeys in the mountains, conveniently numbered 1..N, are once again standing in a row. Monkey i has height H_i (1 <= H_i <= 1,000,000).

Each monkey is looking to his left toward those with higher index numbers. We say that monkey i "looks up" to monkey j if i < j and H_i < H_j. For each monkey i, we would like to know the index of the first monkey in line looked up to by monkey i.

Input

Input consists of several testcases. The format of each case as follow:

Line 1: A single integer: N

Lines 2..N+1: Line i+1 contains the single integer: H_i

Output

For each testcase, output N lines. Line i contains a single integer representing the smallest index of a monkey up to which monkey i looks. If no such monkey exists, print 0.

Sample Input

6326112

Sample Output

330660

题意：问你一个猴子向左看最远能看多远。

分析：搞一个单调递增栈就行了。

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<set>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int INF=0x3f3f3f3f;typedef long long LL;const int maxn=1e5+100;int s[maxn],p[maxn];int a[maxn],n,R[maxn];int main(){    while(~scanf("%d",&n))    {        REPF(i,1,n)          scanf("%d",&a[i]);        int top=0;        for(int i=n;i>=1;i--)        {            while(top>0&&a[i]>=a[s[top]])               --top;            R[i]=s[top];            s[++top]=i;        }        for(int i=1;i<=n;i++)            printf("%d\n",R[i]);    }}