連結：戳這裡
C. Woodcutters time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard 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.
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
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].
題意：
給出n顆樹的高度hi以及在x軸上的坐标xi，伐木工去砍樹，可以使得樹向左倒或者向右倒，如果倒下去會壓到左右兩邊的樹的話，這棵樹就不能砍。問最多能砍多少棵樹。
思路：
設定dp狀态：
dp[i][0] 表示目前第i棵樹向左倒，能獲得的最多的樹的數目
dp[i][1] 表示目前第i棵樹不倒，能獲得的最多的樹的數目
dp[i][2] 表示目前第i棵樹向右倒，能獲得的最多的樹的數目
如果這棵樹能向左倒且第i-1顆樹向右倒或者不倒 dp[i][0]=max(dp[i-1][1],dp[i-1][0])+1;
如果這棵樹向左倒且第i-1棵樹向右倒 dp[i][0]=max(dp[i-1][2]+1,dp[i][0]);
如果這棵樹不倒，則直接繼承上1棵樹的所有狀态
如果這棵樹能向右倒 dp[i][2]=上一棵樹的狀态最大值+1
代碼：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
ll dp[100100][3]; /// 0倒向左邊 1不動 2倒向右邊
ll x[100100],h[100100];
ll L[100100],R[100100];
int Max(int a,int b,int c){
a=max(a,b);
return max(a,c);
}
int n;
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%I64d%I64d",&x[i],&h[i]);
dp[1][0]=1;
if(x[1]+h[1]<x[2]) dp[1][2]=1;
for(int i=2;i<=n;i++){
ll tmp=Max(dp[i-1][0],dp[i-1][1],dp[i-1][2]);
dp[i][0]=dp[i][1]=dp[i][2]=tmp;
if(x[i]-h[i]>x[i-1]) dp[i][0]=Max(dp[i][0],dp[i-1][1]+1,dp[i-1][0]+1);
if(x[i]-x[i-1]>h[i]+h[i-1]) dp[i][0]=Max(dp[i][0],dp[i-1][1]+1,dp[i-1][2]+1);
if(i+1<=n && x[i]+h[i]<x[i+1]) dp[i][2]=tmp+1;
if(i==n) dp[i][2]=tmp+1;
}
printf("%d\n",Max(dp[n][0],dp[n][1],dp[n][2]));
return 0;
}
dp[i][0] 表示目前第i棵樹向左倒，能獲得的最多的樹的數目 dp[i][0] 表示目前第i棵樹向左倒，能獲得的最多的樹的數目