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 6Cow#1 can see the hairstyle of cows #2, 3, 4Cow#2 can see no cow's hairstyleCow#3 can see the hairstyle of cow #4Cow#4 can see no cow's hairstyleCow#5 can see the hairstyle of cow 6Cow#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
Line 1: The number of cows, N. Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.Output
Line 1: A single integer that is the sum of c 1 through cN.Sample Input
6 10 3 7 4 12 2Sample Output
5Source
USACO 2006 November Silver题解：计算每头牛可以看到牛的头数，可以等效于每头牛可以被看到的次数
从左到右依次读取当前牛的高度，从栈顶开始把高度小于或等于当前牛的高度的那些元素删除，此时栈中剩下的元素的数量就是可以看见当前牛的其他牛的数量，求和就是答案
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <stack> #include <string> #include <cmath> #include <cstdio> #include <functional> typedef long long ll; #define N 1005 using namespace std; int main() { int n; while(scanf("%d",&n)==1){ int num; ll sum=0; scanf("%d",&num); stack<int>s; while(!s.empty()) s.pop(); s.push(num); for(int i=1;i<n;i++){ scanf("%d",&num); while(!s.empty()&&num>=s.top()) //当前牛小于栈顶 s.pop(); sum+=s.size(); s.push(num); } printf("%lld\n",sum); } return 0;
转载于:https://www.cnblogs.com/wuhu-JJJ/p/11192414.html
