AtCoder Beginner Contest 103 D - Islands War

Time limit : 2sec / Memory limit : 1024MB

Score : 400 points

Problem Statement

There are N islands lining up from west to east, connected by N−1 bridges.

The i-th bridge connects the i-th island from the west and the (i+1)-th island from the west.

One day, disputes took place between some islands, and there were M requests from the inhabitants of the islands:

Request i: A dispute took place between the ai-th island from the west and the bi-th island from the west. Please make traveling between these islands with bridges impossible.

You decided to remove some bridges to meet all these M requests.

Find the minimum number of bridges that must be removed.

Constraints

• All values in input are integers.
• 2≤N≤105
• 1≤M≤105
• 1≤ai<bi≤N
• All pairs (ai,bi) are distinct.

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Input

Input is given from Standard Input in the following format:

N M
a1 b1
a2 b2
:
aM bM

Output

Print the minimum number of bridges that must be removed.

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Sample Input 1

Copy

5 2
1 4
2 5

Sample Output 1

Copy

1

The requests can be met by removing the bridge connecting the second and third islands from the west.

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Sample Input 2

Copy

9 5
1 8
2 7
3 5
4 6
7 9

Sample Output 2

Copy

2

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Sample Input 3

Copy

5 10
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5

Sample Output 3

Copy

4

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#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
struct node{
	int l,r;
}a[100000+20];
bool cmp(node a,node b)
{
	if(a.r==b.r)
		return a.l<b.l;//右边相同比较左边
	return a.r<b.r;//对数据右排序
}
int main ()
{
	int n,m,i;
	scanf("%d%d",&n,&m);
	for(i=0;i<m;i++)
		scanf("%d %d",&a[i].l,&a[i].r);
	int mn=a[0].r;
	for(i=0;i<m;i++)
		mn=min(a[i].r,mn);
		
	sort(a,a+m,cmp);//排序
int now=mn,ans=1;//找到最小的
	for(i=1;i<m;i++)
	{
		if(a[i].l>=now)
		{
			now=a[i].r;//如果下一个数据的左边大于切断桥的右边则仍然需要切桥
			ans++;
		}

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

切的桥始终在右区间的。