CF 545C Woodcutters

C. Woodcutters

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Sample test(s)

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

Note

In the first sample you can fell the trees like that:

fell the 1-st tree to the left — now it occupies segment [ - 1;1]

fell the 2-nd tree to the right — now it occupies segment [2;3]

leave the 3-rd tree — it occupies point 5

leave the 4-th tree — it occupies point 10

fell the 5-th tree to the right — now it occupies segment [19;20]

In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].

【题目大意】：

有n棵树，给出每棵树的位置和高度，然后把树砍掉，树可以向左倒也可以向右倒。输出最多能砍几棵树。 

【思路】：利用贪心的思想。第一棵树的左边和最后一棵树的右边没树，所以他们向两边倒，然后对于中间的树来说，首先先向左边倒，然后左边距离如果不够的话再向右边倒，向右倒的时候注意更新一下距离。
用last记录左边的端点。应为小心，会有那种情况  两棵树：10 2 和20 2 .这两棵树完全可以倒向同一边。
代码还要注意只有一棵树的情况。

#include<iostream>
using namespace std;
int x[100005];
int h[100005];
int main()
{
int n;
cin >> n;
for(int i=0;i<n;i++)
{
cin >> x[i] >> h[i];
}
int ans = 0;
ans+=2;
int flag = 0;
int last = x[0];
for(int i=1;i<n-1;i++)
{
if(x[i]-last>h[i])
{
ans++;
last = x[i];
continue;
}
if(x[i+1]-x[i]>h[i])
{
ans++;
last = x[i]+h[i];
continue;
}
last = x[i];
}
if(n==1)ans = 1;
cout<<ans<<endl;
return 0;
}