Codeforces Beta Round #4 (Div. 2 Only)D. Mysterious Present-优快云博客

本文链接：https://blog.youkuaiyun.com/tjqACM/article/details/38826147

针对一个有趣的问题，即如何从一系列信封中找到能够形成最长递增链的组合，使得最小的信封能够容纳特定尺寸的卡片。文章通过详细的算法解释和代码实现，展示了如何解决这一问题。

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Peter decided to wish happy birthday to his friend from Australia and send him a card. To make his present more mysterious, he decided to make a chain. Chain here is such a sequence of envelopes A = {a1, a2, ..., an}, where the width and the height of the i-th envelope is strictly higher than the width and the height of the (i - 1)-th envelope respectively. Chain size is the number of envelopes in the chain.

Peter wants to make the chain of the maximum size from the envelopes he has, the chain should be such, that he'll be able to put a card into it. The card fits into the chain if its width and height is lower than the width and the height of the smallest envelope in the chain respectively. It's forbidden to turn the card and the envelopes.

Peter has very many envelopes and very little time, this hard task is entrusted to you.

Input

The first line contains integers n, w, h (1 ≤ n ≤ 5000, 1 ≤ w, h ≤ 106) — amount of envelopes Peter has, the card width and height respectively. Then there follow n lines, each of them contains two integer numbers wi and hi— width and height of the i-th envelope (1 ≤ wi, hi ≤ 106).

Output

In the first line print the maximum chain size. In the second line print the numbers of the envelopes (separated by space), forming the required chain, starting with the number of the smallest envelope. Remember, please, that the card should fit into the smallest envelope. If the chain of maximum size is not unique, print any of the answers.

If the card does not fit into any of the envelopes, print number 0 in the single line.

Sample test(s)

input
2 1 1
2 2
2 2

output
1
1 

input
3 3 3
5 4
12 11
9 8

output
3
1 3 2 


题意：Peter 要给朋友送卡片，卡片宽w高h，他有n个信封，他希望大信封里装小信封，最后装卡片，问你最多用几个信封。


思路：对于信封先按宽排序，最后求满足大于卡片长宽的最长上升列。


代码：
#include<cstdio>
#include<cstring>
#include<stack>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 5005

int dp[N],s[N];
struct node
{
    int x,y,id;
}e[N];
bool cmp(node a,node b)
{
    return a.x<b.x;
}
int main()
{
    int n,w,h,i,j;
    while(scanf("%d%d%d",&n,&w,&h)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&e[i].x,&e[i].y);
            e[i].id=i;
        }
        sort(e+1,e+n+1,cmp);
        memset(dp,0,sizeof(dp));
        memset(s,0,sizeof(s));
        int t;
        int loc;
        for(i=1;i<=n;i++)
        {
            t=0;
            loc=i;
            if(e[i].x<=w||e[i].y<=h)continue;
            for(j=1;j<=i;j++)
            {
                if(e[j].x<=w||e[j].y<=h)continue;
                if(e[j].x!=e[i].x&&e[i].y>e[j].y)
                {
                    if(dp[j]>t)
                    {
                        t=dp[j];
                        loc=j;
                    }
                }
            }
            dp[i]=t+1;
            s[e[i].id]=e[loc].id;
        }
        int ans=-1;
        for(i=1;i<=n;i++)
        {
            if(ans<dp[i])
            {
                ans=dp[i];
                loc=e[i].id;
            }
        }
        stack<int>ss;
        printf("%d\n",ans);
        if(ans!=0)
        {
            while(1)
            {
                ss.push(loc);
                if(s[loc]==loc)break;
                loc=s[loc];
            }
            while(ss.size())
            {
                printf("%d ",ss.top());
                ss.pop();
            }
            printf("\n");
        }
    }
    return 0;
}