POJ3250 - Bad Hair Day - 单调栈

1.题目描述：

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 10300 Accepted: 3458

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 h_i (1 ≤ h_i ≤ 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 c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_N.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 c_N.

Sample Input

6 10 3 7 4 12 2

Sample Output

5 2.题意概述

n个牛排成一列向右看，牛i能看到牛j的头顶，当且仅当牛j在牛i的右边并且牛i与牛j之间的所有牛均比牛i矮。设牛i能看到的牛数为Ci，求∑Ci 3.解题思路：

求所有牛总共能看到多少牛，可以转化为：这n头牛共能被多少头牛看见。 当我们新加入一个高度值时，如果栈中存在元素小于新加入的高度值，那么这些小的牛肯定看不见这个高度的牛(那就看不见这头牛后边的所有牛)， 所以就可以把这些元素弹出。每次加入新元素，并执行完弹出操作后，栈中元素个数便是可以看见这个牛的“牛数”~~~。

4.AC代码：

#include <cstdio> #include <iostream> #include <cstring> #include <string> #include <algorithm> #include <functional> #include <cmath> #include <vector> #include <queue> #include <stack> #include <map> #include <set> #include <ctime> #define INF 0x7fffffff #define maxn 80800 #define eps 1e-6 #define pi acos(-1.0) #define e 2.718281828459 #define mod (int)1e9 + 7; using namespace std; typedef long long ll; int a[maxn]; int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); long _begin_time = clock(); #endif int n; stack<int> s; while (scanf("%d", &n) != EOF) { while (!s.empty()) s.pop(); ll ans = 0; for (int i = 1; i <= n; i++) { int h; scanf("%d", &h); while (!s.empty() && h >= s.top()) s.pop(); ans += 1LL * s.size(); s.push(h); } printf("%lld\n", ans); } #ifndef ONLINE_JUDGE long _end_time = clock(); printf("time = %ld ms\n", _end_time - _begin_time); #endif return 0; }