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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
Output
Sample Input
610374122
Sample Output
5
一些牛从左到右排列。全部的牛都从左往右看，左边的牛仅仅能看到右边的比它身高严格小的牛的发型。假设被一个大于等于它身高的牛挡住。那么它就不能看到再右边的牛。要求每头牛能够看到其它牛的总数。转化一下，事实上就是求每头牛被看到的总次数。能够用单调栈，每次删除栈中比当前牛的身高小于等于的数。事实上这题也能够看做是单调队列。但由于不用对对首操作。所以可看做退化为了栈。
#include#include #include #include #include using namespace std;#define maxn 80600int a[maxn],stack[maxn];int main(){ int n,m,i,j,top; __int64 sum; while(scanf("%d",&n)!=EOF){ memset(stack,0,sizeof(stack)); top=0;sum=0; for(i=1;i<=n;i++){ scanf("%d",&a[i]); while(top>0 && a[i]>=stack[top])top--; sum+=top; stack[++top]=a[i]; } printf("%I64d\n",sum); } return 0;}