标签：limit therefore turn pst max input his main cst
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 21084 | Accepted: 7202 |
Description
Some of Farmer John‘s N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows‘ heads.
Each cow i has a specified height hi (1 ≤ hi ≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.
Consider this example:
=
= =
= - = Cows facing right -->
= = =
= - = = =
= = = = = =
1 2 3 4 5 6
Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow‘s hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow‘s hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!
Let ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.
Input
Output
Sample Input
6 10 3 7 4 12 2
Sample Output
5
题意：个子高的牛能看到个子低的牛，且每头牛的视线都是一个方向,问每头牛分别能看到视线方向上的多少头牛。
思路：单调栈,从后往前考虑队列,若当前的牛能看到存储在栈中的牛，则把栈中的牛从栈中抛出,直到栈为空或者栈中的牛的身高大于等于当前牛的身高为止，将当前的牛压入栈中，每次执行这样的操作即可。
AC代码：
#define _CRT_SECURE_NO_DEPRECATE #include<iostream> #include<algorithm> #include<vector> #include<cstring> #include<string> #include<cmath> using namespace std; #define INF 0x3f3f3f3f #define N_MAX 80000+20 typedef long long ll; int n,h[N_MAX]; int st[N_MAX],num[N_MAX]; int main() { scanf("%d",&n); for (int i = 0; i < n; i++)scanf("%d",&h[i]); int t = 0;//栈的大小 for (int i = n-1; i >=0; i--) { while (t > 0 && h[st[t - 1]] <h[i])t--; num[i] = t == 0 ? n - 1 - i : st[t - 1] - i - 1; st[t++]= i; } ll sum = 0; for (int i = 0; i < n; i++) sum += num[i]; printf("%lld\n",sum); return 0; }
标签：limit therefore turn pst max input his main cst
原文地址：https://www.cnblogs.com/ZefengYao/p/9021025.html
