Poj 3250 Bad Hair Day 【单调栈】

题目链接：Poj 3250 Bad Hair Day

Bad Hair Day
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 16700 Accepted: 5621
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

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

题意：给定n头牛的高度，设d(i)为第i头牛可以看到(向右看)的其它牛的总数。求∑ni=1d[i]

思路：反过来想，求第i头牛可以被左边多少头牛看到。维护一个单调栈即可。

AC代码：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#define CLR(a, b) memset(a, (b), sizeof(a))using namespace std;typedef long long LL;const int MAXN = 1e4 +10;const int INF = 0x3f3f3f3f;int Stack[80000+1];int main(){    int N;    while(scanf("%d", &N) != EOF) {        LL ans = 0; int top = 0;        for(int i = 1; i <= N; i++) {            int a; scanf("%d", &a);            while(top && Stack[top-1] <= a) {                top--;            }            ans += top;            Stack[top++] = a;        }        printf("%lld\n", ans);    }    return 0;}