poj 3250 Bad Hair Day (单调栈)
2015-12-15 10:37
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Bad Hair Day
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:
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
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 c1 through cN.
Sample Input
Sample Output
单调栈的入门题:单调栈就是维护一个栈,栈中的元素都保持着单调递增或递减的顺序。
题目意思:有n仅仅牛站在一排,给出队伍中每仅仅牛的高度。每仅仅牛仅仅能看到它右边比它矮的牛。求全部的牛能看到的牛数之和。
当我们新增加一个高度值时,假设栈中存在元素小于新增加的高度值。那这个牛肯定看不见这个高度的牛,就把这个元素弹栈。每次增加新元素,并运行完弹出操作后,栈中元素个数便是能够看见这个牛的“牛数”。
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 14883 | Accepted: 4940 |
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
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 c1 through cN.
Sample Input
6 10 3 7 4 12 2
Sample Output
5
单调栈的入门题:单调栈就是维护一个栈,栈中的元素都保持着单调递增或递减的顺序。
题目意思:有n仅仅牛站在一排,给出队伍中每仅仅牛的高度。每仅仅牛仅仅能看到它右边比它矮的牛。求全部的牛能看到的牛数之和。
当我们新增加一个高度值时,假设栈中存在元素小于新增加的高度值。那这个牛肯定看不见这个高度的牛,就把这个元素弹栈。每次增加新元素,并运行完弹出操作后,栈中元素个数便是能够看见这个牛的“牛数”。
#include <iostream> #include <cstdio> #include <cstring> #include <stack> typedef long long ll; using namespace std; int main() { int n; ll heigh,ans; stack<ll>s; while(scanf("%d",&n)!=EOF) { ans=0; cin>>heigh; s.push(heigh);//入栈 for(int i=1;i<n;i++) { cin>>heigh; while(!s.empty()&&s.top()<=heigh) //比較栈顶元素和新增加元素的关系 { s.pop(); } ans+=s.size(); s.push(heigh); } cout<<ans<<endl; while(!s.empty()) s.pop();//空栈 } return 0; }
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