ACdream 1216——Beautiful People——————【二维LIS，nlogn处理】

Beautiful People

Special Judge Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

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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 — Si and Bi 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

4
1 1
1 2
2 1
2 2

Sample Output

2
1 4

Source

Andrew Stankevich Contest 1

 

 

题目大意：给你n个人，每个人有两个能力ai，bi。如果两个人的ai>=aj&&bi<=bj或ai<=aj&&bi>=bj的时候，他们两个会有仇恨。现在给你n个人的能力值，问你最多能邀请多少个人去聚会，且任意两人之间没有仇恨。

 

解题思路：要使每个人之间都没有仇恨的话，那么保证ai<aj&&bi<bj即可。那么我们可以首先将a能力排序，从小到大，如果a能力相同的时候，我们保证b能力大的排在前面，不然在endminv数组中的值会发生错误替换，模拟下面给的样例就知道为什么要排序b能力了。然后只需要在b能力上进行LIS操即可。这里用的是nlogn的LIS写法。LCS也有nlogn的写法，其实就是转化成求LIS的。

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
using namespace std;
const int maxn=1e5+200;
const int INF = 0x3f3f3f3f;
int dp[maxn],endminv[maxn];	//dp[i]表示以i结尾的最长递增子序列长度		//endminv[i]表示LIS长度为i时，LIS结尾的最小值是多少
int path[maxn];
struct Number{
    int x,y,idx;
}numbers[maxn];
bool cmp(Number a,Number b){
    if(a.x==b.x)
    return a.y>b.y;     //必须有这个排序，下面那个样例就卡这里不排序的写法
    return a.x<b.x;
}
int BinSearch(int l,int r,int key){	//二分查找大于等于key的第一个位置（下界）
    int md;
    while(l<r){
        md=(l+r)/2;
        if(endminv[md]>key){
            r=md;
        }else if(endminv[md]<key){
            l=md+1;
        }else{
            return md;
        }
    }
    return l;
}
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        for(int i=1;i<=n;i++){
            scanf("%d%d",&numbers[i].x,&numbers[i].y);
            numbers[i].idx=i;
        }
        sort(numbers+1,numbers+1+n,cmp);
        memset(endminv,INF,sizeof(endminv));
        int len=1,x,maxv=0;
        for(int i=1;i<=n;i++){
            x=BinSearch(1,n,numbers[i].y);
            endminv[x]=numbers[i].y;
            if(x>=len){
                len++;
            }
            dp[i]=x;
        }
        len = len -1;
        printf("%d
",len);
        int i,minx=INF,miny=INF;
        int fir=1;
        for(i=n;i>=1;i--){
            if(dp[i]==len){
                if(numbers[i].x<minx&&numbers[i].y<miny){
                    len--;
                    minx=numbers[i].x;
                    miny=numbers[i].y;
                    if(fir){
                        fir=0;
                    }else printf(" ");
                    printf("%d",numbers[i].idx);
                }
            }
        }
    }
    return 0;
}

/*
5
1 3
2 6
3 1
3 2
3 3

*/