codeforces 528B Clique Problem （贪心 + 思维）

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#codeforces #题解 #贪心 #思维

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探讨了在特定条件下的最大团问题，即在由坐标轴上的点构成的图中寻找最大团。通过将问题转化为寻找数轴上最多不重叠的线段数量，使用贪心算法解决。

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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 graph G, 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.

Examples
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.

在这里插入图片描述

题目链接

http://codeforces.com/contest/528/problem/B

题目大意

在坐标轴上有N个点，每个点权重为w。若满足两点距离大于两点权重之和，则这两点可以连接起来。问已知的N个点中最多有多少点满足两两相连

数据范围

n (1 ≤ n ≤ 200 000)
xi, wi (0 ≤ xi ≤ 109, 1 ≤ wi ≤ 109)

解题思路

已知|x1 - x2|>w1 + w2;
则当x1 > x2时，x1 - w1 > x2 + w2;
当x2 > x1时，x2 - w2 > x1 + w1;
此题可转化为将每点变为长度为2w的线段（x - w,x + w），判断数轴上最多能存在多少不重叠的线段，即为本题答案。此时用贪心，先x + w小的优先。
类似“今年暑假不AC”

解决代码

#include<cstdio>
#include<algorithm>
using namespace std;
struct A{
int l,r;
}a[200005];
bool cmp(A x,A y){
return x.r < y.r;
}
int main()
{
int n,x,w,ans = 0,t = -1000000005;
scanf("%d",&n);
for(int i = 1;i <= n;i++){
	scanf("%d %d",&x,&w);
	a[i].l = x - w;
	a[i].r = x + w;
}
sort(a + 1,a + n + 1,cmp);

for(int i = 1;i <= n;i++){
	if(a[i].l >= t) {
		t = a[i].r;
		ans++;
		}
	}
	printf("%d\n",ans);
return 0;
}