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
1 #include <stdio.h>
2 #include <string.h>
3 #include <iostream>
4 #include <string>
5 #include <math.h>
6 #include <algorithm>
7 #include <vector>
8 #include <queue>
9 #include <set>
10 #include <map>
11 #include <math.h>
12 const int INF=0x3f3f3f3f;
13 typedef long long LL;
14 const int mod=1e9+7;
15 //const int mod=1e8;
16 //const double PI=acos(-1);
17 const int maxn=1e5+10;
18 using namespace std;
19 //ios::sync_with_stdio(false);
20 // cin.tie(NULL);
21
22 int A[maxn];
23 int ans[maxn];
24
25 int main()
26 {
27 int n;
28 scanf("%d",&n);
29 for (int i=1;i<=n;i++)
30 scanf("%d",&A[i]);
31 vector<int> vt;
32 vt.push_back(1);
33 int MIN=A[1];
34 for (int i = 2; i <= n; i++)
35 {
36 if(A[i]>=MIN)
37 {
38 while (!vt.empty())
39 {
40 if (A[*(vt.end()-1)]<A[i])
41 {
42 ans[*(vt.end()-1)]=i;
43 vt.erase(vt.end()-1);
44 }
45 else
46 break;
47 }
48 vt.push_back(i);
49 }
50 else
51 {
52 vt.push_back(i);
53 MIN=ans[i];
54 }
55 }
56 for (vector<int>::iterator it=vt.begin();it!=vt.end();it++)
57 {
58 ans[*it]=0;
59 }
60 for (int i=1;i<=n;i++)
61 printf("%d\n",ans[i]);
62 return 0;
63 }
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
1 #include <stdio.h>
2 #include <string.h>
3 #include <iostream>
4 #include <string>
5 #include <math.h>
6 #include <algorithm>
7 #include <vector>
8 #include <stack>
9 #include <queue>
10 #include <set>
11 #include <map>
12 #include <math.h>
13 const int INF=0x3f3f3f3f;
14 typedef long long LL;
15 const int mod=1e9+7;
16 const int maxn=1e5+10;
17 using namespace std;
18
19 int A[maxn];
20 int ans[maxn];
21 stack<int> sk;
22
23 int main()
24 {
25 int n;
26 scanf("%d",&n);
27 for (int i=1;i<=n;i++)
28 scanf("%d",&A[i]);
29 for(int i=n;i>=1;i--)
30 {
31 while(!sk.empty()&&A[sk.top()]<=A[i])
32 sk.pop();
33 if(!sk.empty())
34 ans[i]=sk.top();
35 else
36 ans[i]=0;
37 sk.push(i);
38 }
39 for (int i=1;i<=n;i++)
40 printf("%d\n",ans[i]);
41 return 0;
42 }
原文地址：https://www.cnblogs.com/jiamian/p/11495076.html