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.
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.
Print a single number — the maximum number of trees that you can cut down by the given rules.
5
1 2
2 1
5 10
10 9
19 1
3
5
1 2
2 1
5 10
10 9
20 1
4
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].
题目大意是给出树的点和他们的高度，伐木工砍伐时可以让他们向左倒或向右倒，倒下后占据覆盖的坐标，但不能倒到已经被占据的点上。
是一道简单的dp和贪心题，可以看到，第一棵树和最后一棵树都是必定可以砍倒的，而从左到右遍历中间的树尽量让他们向左倒，这样倒下后占据的点不会影响到后面的点，每棵树给它一个occupy属性表明它占据的坐标的最大值。dp[i]=max(dp[i-1],i向左倒,i向右倒)。
AC代码：
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 }