NYOJ练习题 how many hairstyles can they see?
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). 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.
For example:
There are six cows, heights are 10 2 7 3 12 2.
Now 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!
610 2 7 3 12 2
5
题意：有n头牛面向右排队，每头牛只能看到右边比它矮的牛，右边比它高的牛会挡住视线；如果比它高的牛的右边还有比它矮的牛，它同样看不到那些牛。求这n头牛能看到的其他牛的数量的总和。
解法：使用栈。如果栈顶的数大于当前元素，说明栈顶的牛可以看见当前的牛，则把当前的元素压入栈中；否则，说明栈顶这头牛看不到当前位置之后的牛了，就可以计算出栈顶的牛可以看到的牛的个数，并让栈顶元素出栈，然后继续比较栈顶元素，直到栈顶元素大于当前位置，把当前元素压入栈中。
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 8e4 + 10;int sta[N], pos[N];int main(){ int n, i, a; while(~scanf("%d",&n)) { int top = 0; scanf("%d",&a); sta[++top] = a; pos[top] = 1; int ans = 0; for(i = 2; i <= n; i++) { scanf("%d",&a); while(top > 0 && sta[top] <= a) { ans += i - pos[top] - 1; //求出的ans是top位置的牛能看到的牛的数量 top--; } sta[++top] = a; pos[top] = i; } top--; //最后一头牛一头也看不到 while(top > 0) { ans += n - pos[top]; top--; } printf("%d\n",ans); } return 0;}/*当找到一个比栈顶元素大的数时，就让栈顶元素出栈，直到栈空，把当前元素压入栈中*/