CodeForces - 545C Woodcutters dp

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Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.

There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of trees.

Next n lines contain pairs of integers xi, hi (1 ≤ xi, hi ≤ 109) — the coordinate and the height of the і-th tree.

The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.

Output

Print a single number — the maximum number of trees that you can cut down by the given rules.

Examples

Input

5
1 2
2 1
5 10
10 9
19 1

Output

3

Input

5
1 2
2 1
5 10
10 9
20 1

Output

4

题意：有n棵树，最多能砍到多少树，砍倒的树不能碰到其他树，可以向右倒也可向左倒

题解：dp[i][0] 表示 第i棵树，不砍

           dp[i][1] 表示 第i棵树  向左倒   判断 是否可以向左倒，是否可以i-1右倒和i左倒

           dp[i][2] 表示 第i棵树  向右倒   判断 是否可以右倒

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+10;
ll dp[N][4],pos[N],h[N];
int n,x;
int main()
{
	pos[0]=-2e9;
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%lld%lld",&pos[i],&h[i]);
	pos[n+1] = 4e9;
	for(int i=1;i<=n;i++)
	{		
		dp[i][0]=max(dp[i-1][0],max(dp[i-1][1],dp[i-1][2]));
		
		if(pos[i]-pos[i-1]>h[i]+h[i-1]) dp[i][1]=max(dp[i-1][0],max(dp[i-1][1],dp[i-1][2])) + 1;
		else if(pos[i]-pos[i-1]>h[i]) dp[i][1]=max(dp[i-1][0],dp[i-1][1]) + 1;
		
		if(pos[i+1]-pos[i]>h[i]) dp[i][2]=max(dp[i-1][0],max(dp[i-1][1],dp[i-1][2])) + 1;

	}
	cout<<max(dp[n][0],max(dp[n][1],dp[n][2]))<<endl;
	return 0;
}