[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

输出

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+;
 const int maxn=1e5+;
 using namespace std;

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

 int main()
 {
     int n;
     scanf("%d",&n);
     for (int i=;i<=n;i++)
         scanf("%d",&A[i]);
     vector<int> vt;
     vt.push_back();
     int MIN=A[];
     for (int i = ; i <= n; i++)
     {
         if(A[i]>=MIN)
         {
             while (!vt.empty())
             {
                 if (A[*(vt.end()-)]<A[i])
                 {
                     ans[*(vt.end()-)]=i;
                     vt.erase(vt.end()-);
                 }
                 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]=;
     }
     for (int i=;i<=n;i++)
         printf("%d\n",ans[i]);
     return ;
 }

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+;
 const int maxn=1e5+;
 using namespace std;

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

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