CF R303 div2 C. 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].
제목은 나무의 점과 높이를 제시하는 것이다. 벌목꾼은 벌목할 때 왼쪽으로 쓰러지거나 오른쪽으로 쓰러지게 할 수 있다. 쓰러지면 덮인 좌표를 차지하지만 이미 점거된 점에 쓰러지면 안 된다.
간단한 dp와 욕심 문제입니다. 첫 번째 나무와 마지막 나무는 반드시 베어 넘어뜨릴 수 있습니다. 왼쪽에서 오른쪽으로 가운데를 두루 돌아다니는 나무는 최대한 왼쪽으로 넘어뜨릴 수 있습니다. 이렇게 쓰러진 후에 차지하는 점은 뒤에 있는 점에 영향을 주지 않습니다. 모든 나무는 occupy 속성을 주어 자신이 차지하는 좌표의 최대치를 나타냅니다.dp[i]=max(dp[i-1], i는 왼쪽으로 넘어지고 i는 오른쪽으로 넘어진다).
AC 코드:

 1 #include<iostream>

 2 #include<cstring>

 3 #include<string>

 4 using namespace std;

 5 int dp[100005];

 6 struct tree{

 7     int h,x,occupy;

 8 }t[100005];

 9 int main(){

10     int n,i;

11     while(cin>>n){

12         memset(dp,0,sizeof(dp));

13         for(i=0;i<n;i++){

14              cin>>t[i].x>>t[i].h;

15              t[i].occupy=t[i].x;

16         } 

17         dp[0]=1;

18         if(n==1){

19             cout<<1<<endl;

20             continue;

21         }

22         if(n==2){

23             cout<<2<<endl;

24             continue;

25         }

26         for(i=1;i<n-1;i++){

27             int th=t[i].h,tx=t[i].x,toc=t[i-1].occupy;

28             int l=dp[i-1],r=dp[i-1];

29             if((tx+th)<t[i+1].x){

30                 r++;

31                 t[i].occupy=tx+th;

32             }

33             if(tx-th>toc){

34                 l++;

35                 t[i].occupy=tx;

36             }

37             dp[i]=max(max(dp[i-1],l),r);

38         }

39         dp[n-1]=dp[n-2]+1;

40         cout<<dp[n-1]<<endl;

41     }

42     return 0;

43 }