51nod 1289 Feeding Frenzy

1289 Feeding Frenzy

There are n fish each fish's position and size are different, they swim along the x-axis, some to the left, some to the right. Swimming speed is the same, two fish meet big fish will eat small fish. The size of each fish and the direction of the swim are given from left to right (0 means left and 1 for right). Ask how many fish you have left after a long enough time. Input

Line 1th: 1 number N, indicating the number of fish (1 <= n <= 100000).
2-n + 1 lines: Two numbers per line a[i], B[i], separated by a space, respectively, the size of the fish and the direction of the swimming (1 <= a[i] <= 10^9,b[i] = 0 or 1,0 means left, 1 to the right).

Output

Output 1 numbers, indicating the number of fish that are eventually left.

Input example

5
4 0
3 1
2 0
1 0
5 0

Output example

2

The fish moving to the right moves into the stack, moving to the left and comparing the elements in the stack until the stack is empty or the fish is eaten.

#include <iostream>
#include <stack>
using namespace std;
struct fish
{
	int weight;
	int direction;
} A[100005];
int main ()
{
	int i,n;
	stack<fish>s;
	cin>>n;
	int ans=n;
	for (i=0; i<n; i++)
	{
		cin>>a[i].weight>>a[i].direction;
	}
	for (i=0; i<n; i++)
	{
		if (a[i].direction==1)
		{
			s.push (a[i]);
		}
		else
		{
			while (!s.empty ())
			{
				if (a[i].weight>s.top (). Weight)
				{
					s.pop (); ans--;
				}
				else
				{
					ans--;	Break
   ;
	}}}} cout<<ans<<endl;
}