POJ 3250:Bad Hair Day 好玩的单调栈
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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
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
Output
Sample Input
610374122
Sample Output
5
题意是站着一排牛,牛从左往右看能看到比自己身高小的牛的发型,但是如果碰到牛的身高比自己高了,那么到此为止。
转换一下思维,想象一下牛是从右往左看只能看到比自己身高大的牛,然后如果身高变小,那么到此为止。
很好玩的题目,用单调栈来做,从左往右读,栈内元素从栈底到栈顶是递增的,这样如果遇到元素比栈顶元素大,那么弹出。然后计算此时栈内元素的个数,相加即得到结果。
代码:
#include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <string> #include <cstring> #pragma warning(disable:4996) using namespace std;#define N 80002long long a[N], stack[N], top;int main(){//freopen("i.txt", "r", stdin);//freopen("o.txt", "w", stdout);long long ans, tmp;int i, j, n;scanf("%d", &n);for (i = 1; i <= n; i++){scanf("%lld", a + i);}a[++n] = 1e9 + 7;top = 0;ans = 0;for (i = 1; i <= n; i++){while (top >= 1 && a[i] >= a[stack[top - 1]]){--top;}ans = ans + top;if (top == 0 || a[i] < a[stack[top - 1]]){stack[top++] = i;}}printf("%lld\n", ans);//system("pause"); return 0;}
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