codforeces 527D 贪心

D. Clique Problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as follows. Consider an undirected graph G. It is required to find a subset of vertices C of the maximum size such that any two of them are connected by an edge in graph G. Sounds simple, doesn't it? Nobody yet knows an algorithm that finds a solution to this problem in polynomial time of the size of the graph. However, as with many other NP-complete problems, the clique problem is easier if you consider a specific type of a graph.

Consider n distinct points on a line. Let the i-th point have the coordinate xi and weight wi. Let's form graphG, whose vertices are these points and edges connect exactly the pairs of points (i, j), such that the distance between them is not less than the sum of their weights, or more formally: |xi - xj| ≥ wi + wj.

Find the size of the maximum clique in such graph.

Input

The first line contains the integer n (1 ≤ n ≤ 200 000) — the number of points.

Each of the next n lines contains two numbers xi, wi (0 ≤ xi ≤ 109, 1 ≤ wi ≤ 109) — the coordinate and the weight of a point. All xi are different.

Output

Print a single number — the number of vertexes in the maximum clique of the given graph.

Sample test(s)

input

4
2 3
3 1
6 1
0 2

output

3

Note

If you happen to know how to solve this problem without using the specific properties of the graph formulated in the problem statement, then you are able to get a prize of one million dollars!

The picture for the sample test.

满足|xi-xj| >= wi+wj  -> xi-wi >= xj+wj则i，j连接一条边。 将所有的数据xi，wi转换成(xi-wi,xi+wi)对。第i个点和前面的点连接时，只需要考虑前xj+wj的最大值就可以了，若大于xj+wj，则前面与j相连的也可以与i相连。把(xi-wi,xi+wi)看成区间的话，求最大的团，其实就是求最多的互不覆盖的区间个数，因此按xi+wi排序，贪心的选择就行了。线性复杂度。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>

using namespace std;

#define maxn 200005
#define inf 0x3f3f3f3f

typedef pair<int,int> p;

p a[maxn];
int n;

int cmp (const p &a , const p &b)
{
    return a.second < b.second;
}

int main()
{
    while(~scanf("%d", &n)){
        int x,w;
        for(int i = 0; i < n; i++){
            scanf("%d%d", &x, &w);
            a[i] = p(x-w,x+w);
        }

        sort(a, a+n,cmp);
        int pre = -inf;
        int ans = 0;
        for(int i = 0; i < n; i++){
            if(a[i].first >= pre)
                ans++, pre = a[i].second;
        }

        printf("%d\n",ans);
    }
    return 0;
}