标签：
http://poj.org/problem?id=3250
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 15985 | Accepted: 5404 |
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
题目大意：
n个数，从左到右排成一排，问每个数的右边比他小的数有几个，然后求和
可以反过来想，看每个数他的左边有多少个数比他大的有几个，然后再加起来
用栈来模拟，先将第一个数压入栈：
如果栈首元素S.top() > a[i],
则栈里面的元素都比a[i]大，那么比a[i]大的数的个数就是栈里面的元素个数S.size();
否则栈首元素S.top() <= a[i],说明S.top()不符合条件（不大于a[i]）
则让栈首元素出栈，继续比较栈内元素如果不大于a[i],就让他出栈
也就是说栈内的元素都是比a[i]大的数
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<stack> #include<algorithm> using namespace std; const int N = 80010; typedef __int64 ll; int a[N]; int main() { stack<int>S; int n; while(~scanf("%d", &n)) { ll sum = 0; for(int i = 0 ; i < n ; i++) scanf("%d", &a[i]); S.push(a[0]);//第一个数进栈 for(int i = 1 ; i < n ; i++)//遍历 { while(!S.empty() && S.top() <= a[i]) S.pop();//出栈 sum += S.size(); S.push(a[i]);//入栈 } printf("%I64d\n", sum); } return 0; }
标签：
原文地址：http://www.cnblogs.com/qq2424260747/p/4977821.html
