[USACO09MAR]向右看齐Look Up（单调栈、在线处理）

https://www.luogu.org/problem/P2947

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 Hi (1 <= Hi <= 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 Hi < Hj. 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.

输入描述:

* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains the single integer: Hi

输出描述:

* 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.

示例1

输入

6 
3 
2 
6 
1 
1 
2 

输出

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.

找每个数右边第一个大于等于它的位置，并输出该位置，如果没有就输出0

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <math.h>
const int INF=0x3f3f3f3f;
typedef long long LL;
const int mod=1e9+7;
const int maxn=1e5+10;
using namespace std;

int A[maxn]; 
int ans[maxn];

int main()
{
    int n;
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        scanf("%d",&A[i]);
    vector<int> vt; 
    vt.push_back(1);
    int MIN=A[1];
    for (int i = 2; i <= n; i++)
    {
        if(A[i]>=MIN) 
        {
            while (!vt.empty()) 
            {
                if (A[*(vt.end()-1)]<A[i])
                {
                    ans[*(vt.end()-1)]=i;
                    vt.erase(vt.end()-1);
                }
                else
                    break;
            }
            vt.push_back(i);
        }
        else
        {
            vt.push_back(i);
            MIN=ans[i];
        }
    } 
    for (vector<int>::iterator it=vt.begin();it!=vt.end();it++) 
    {
        ans[*it]=0;
    }
    for (int i=1;i<=n;i++)
        printf("%d
",ans[i]);
    return 0;
}

其实上面的代码就用到了单调栈的思想

何为单调栈？
解释一下：
单调栈类似单调队列，这道题中要运用到单调栈。
如下：
令f[i]为向右看齐的人的标号
6 3 2 6 1 1 2
f分别为 3 3 0 6 6 0
首先，最后一个人必然没有向右看齐的人的编号
先将最右边的人加入栈
接着，我们发现1，1比2小，先加入栈，当前f值为6
接着，又来了一个1，发现1=1，弹出1，接着发现1<2,则将1加进栈，当前f值为6
接着，来了一个6，6>2,弹出2，当前f值为0
接着，来了一个2，2<6，加入2，当前f值为3
最后，来了一个3，3>2，弹出2，将3加入栈，当前f值为3
最后的栈中有3和6
最后的答案就是3 3 0 6 6 0
每次只在栈中存数字标号，进来一个数，就把小于等于他的数全部弹出，若剩下有数，则答案是剩下的数，否则答案是0
为什么？
因为：
若x小于当前数，x不会成为答案。
这就是单调栈的用法

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <math.h>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <math.h>
const int INF=0x3f3f3f3f;
typedef long long LL;
const int mod=1e9+7;
const int maxn=1e5+10;
using namespace std;

int A[maxn]; 
int ans[maxn];
stack<int> sk; 

int main()
{
    int n;
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        scanf("%d",&A[i]);
    for(int i=n;i>=1;i--)
    {
        while(!sk.empty()&&A[sk.top()]<=A[i])
            sk.pop();
        if(!sk.empty())
            ans[i]=sk.top();
        else
            ans[i]=0;
        sk.push(i);
    }
    for (int i=1;i<=n;i++)
        printf("%d
",ans[i]);
    return 0;
}