SGU 199 Beautiful People lcs O(nlogn)算法

SGU 199 Beautiful People lcs O(nlogn)算法

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199. Beautiful People

time limit per test: 0.25 sec.
memory limit per test: 65536 KB input: standard
output: standard


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 S_i and beauty B_i . 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 S_i ≤ S_j and B_i ≥ B_j or if S_i ≥ S_j and B_i ≤ B_j (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 presti≥ 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 ≤ S_i, B_i ≤ 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 test(s)
Input

4 1 1 1 2 2 1 2 2 Output

2
1 4

數據有點大

需要用到O(nlogn)的算法

二分= =

這題的狀態需要注意一下

因為需要輸出路徑

所以不能單純記錄那個值

而是應該記錄這個值的位置

#include
#include
#include
using namespace std;
const int MAXN=100100;
int id[MAXN],pre[MAXN];
struct Node
{
    int a,b,id;
}node[100100];
int n;
int cmp(Node x,Node y)
{
    if(x.a!=y.a)
        return x.ay.b;
}
void print(int x)
{
    if(x==-1)
        return ;
    print(pre[x]);
    printf("%d ",node[x].id);
}
int main()
{
    int i,left,right,mid;
    int len;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d %d",&node[i].a,&node[i].b);
            node[i].id=i;
        }
        sort(node+1,node+1+n,cmp);
        //for(i=1;i<=n;i++)
          //  printf("%d %d %d\n",node[i].a,node[i].b,node[i].id);
        len=1;
        memset(pre,-1,sizeof(pre));
        id[0]=1;
        for(i=2;i<=n;i++)
        {
            if(node[i].b>node[id[len-1]].b)
            {
                id[len++]=i;
                pre[i]=id[len-2];
                continue;
            }
            left=0;
            right=len-1;
            while(left