Beautiful People

最新推荐文章于 2019-08-18 09:31:05 发布

梁千仞

最新推荐文章于 2019-08-18 09:31:05 发布

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在一个城市中最负盛名的体育俱乐部中，每位成员都有自己的力量和美丽属性，并且成员间存在互相厌恶的关系。为了庆祝新年，俱乐部计划组织一场派对，但为了避免相互厌恶的成员发生冲突，需要找出能被邀请参加派对的最大数量的成员。此问题可通过编程解决。

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Beautiful People

Time Limit: 10000/5000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) Special Judge

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Problem Description

The most prestigious sports club in one city has exactly N members. Each of its members is strong and beautiful. More precisely, i-th member of this club (members being numbered by the time they entered the club) has strength Si and beauty Bi. Since this is a very prestigious club, its members are very rich and therefore extraordinary people, so they often extremely hate each other. Strictly speaking, i-th member of the club Mr X hates j-th member of the club Mr Y if Si <= Sj and Bi >= Bj or if Si >= Sj and Bi <= Bj (if both properties of Mr X are greater then corresponding properties of Mr Y, he doesn't even notice him, on the other hand, if both of his properties are less, he respects Mr Y very much).

To celebrate a new 2003 year, the administration of the club is planning to organize a party. However they are afraid that if two people who hate each other would simultaneouly attend the party, after a drink or two they would start a fight. So no two people who hate each other should be invited. On the other hand, to keep the club prestige at the apropriate level, administration wants to invite as many people as possible.

Being the only one among administration who is not afraid of touching a computer, you are to write a program which would find out whom to invite to the party.

Input

The first line of the input file contains integer N — the number of members of the club. ( 2 ≤ N ≤ 100 000). Next N lines contain two numbers each — S i and B i respectively ( 1 ≤ Si, Bi ≤ 109).

Output

On the first line of the output file print the maximum number of the people that can be invited to the party. On the second line output N integers — numbers of members to be invited in arbitrary order. If several solutions exist, output any one.

Sample Input

Sample Output

2
1 4

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=100001;
struct P{
	int s,b,id;
}p[N];

bool cmp(P a,P b)
{
	if(a.s==b.s)	return a.b>b.b;
	return a.s<b.s;
}

int dp[N];
int pre[N];
int main()
{
	int T,n,i,k,l,r,m;
		scanf("%d",&n);
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&p[i].s,&p[i].b);
			p[i].id=i;
		}	
		sort(p+1,p+n+1,cmp);
		dp[1] = 1,k = 1;
		for(i=2;i<=n;i++)
		{
			if(p[i].b > p[dp[k]].b)
			{
				pre[i] = dp[k];
				dp[++k] = i;
			}
			else{
				l=1,r=k;
				while(l<r)
				{
					m=(l+r)>>1;
					if(p[dp[m]].b<p[i].b)l=m+1;
					else r=m;
				}
				
				dp[l] = i;
				pre[i] = dp[l-1];
			}
		}
		printf("%d\n",k);
		i = dp[k--];
		printf("%d",p[i].id);
		while(k--){
			i = pre[i];
			printf(" %d",p[i].id);
		}
		puts("");
	return 0;
}

心好累啊，完全不明白什么意思，这代码