hdu5360||多校联合第6场1008 贪心

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概述

http://acm.hdu.edu.cn/showproblem.php?pid=5360

Problem Description

There are

n soda conveniently labeled by

1,2,…,n . beta, their best friends, wants to invite some soda to go hiking. The

i -th soda will go hiking if the total number of soda that go hiking except him is no less than

li and no larger than

ri . beta will follow the rules below to invite soda one by one:
1. he selects a soda not invited before;
2. he tells soda the number of soda who agree to go hiking by now;
3. soda will agree or disagree according to the number he hears.

Note: beta will always tell the truth and soda will agree if and only if the number he hears is no less than

li and no larger than

ri , otherwise he will disagree. Once soda agrees to go hiking he will not regret even if the final total number fails to meet some soda's will.

Help beta design an invitation order that the number of soda who agree to go hiking is maximum.

Input

There are multiple test cases. The first line of input contains an integer

T , indicating the number of test cases. For each test case:

The first contains an integer

(1≤n≤105) , the number of soda. The second line constains

n integers

l1,l2,…,ln . The third line constains

n integers

r1,r2,…,rn .

(0≤li≤ri≤n)
It is guaranteed that the total number of soda in the input doesn't exceed 1000000. The number of test cases in the input doesn't exceed 600.

Output

For each test case, output the maximum number of soda. Then in the second line output a permutation of

1,2,…,n denoting the invitation order. If there are multiple solutions, print any of them.

Sample Input

  
  
   
   4
8
4 1 3 2 2 1 0 3
5 3 6 4 2 1 7 6
8
3 3 2 0 5 0 3 6
4 5 2 7 7 6 7 6
8
2 2 3 3 3 0 0 2
7 4 3 6 3 2 2 5
8
5 6 5 3 3 1 2 4
6 7 7 6 5 4 3 5

Sample Output

  
  
   
   7
1 7 6 5 2 4 3 8
8
4 6 3 1 2 5 8 7
7
3 6 7 1 5 2 8 4
0
1 2 3 4 5 6 7 8

/**
hdu5360||多校联合第6场1008  贪心
题目大意：xxx要邀请n个人去玩，对于第i个人如果已经邀请的人数在(li,ri)之间，他就会去。问用怎样的邀请顺序能邀请到最多的人？
解题思路：和去年上海亚洲赛的一道题挺像。我的想法是先将n个人以li递增排个序，然后从前往后遍历，当前已经邀请的为x人，在所有
          li>=x的的人中取ri最小(set维护)的即可。复杂度O(nlogn)
*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<set>
using namespace std;
const int maxn=100005;

struct note
{
    int l,r,id;
    bool operator <(const note &other)const
    {
        return l<other.l;
    }
}a[maxn];

int n,num[maxn],flag[maxn];
set<pair<int,int> >st;

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i].l);
            a[i].id=i+1;
        }
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i].r);
        }
        sort(a,a+n);
        int x=0;
        memset(flag,0,sizeof(flag));
        for(int i=0;;)
        {
            while(x>=a[i].l&&i<n)
            {
                st.insert(make_pair(a[i].r,a[i].id));
                i++;
            }
            while(((*st.begin()).first<x)&&st.size()>0)st.erase(st.begin());
            if(st.size()==0)break;
            int cnt=(*st.begin()).second;
            num[x++]=cnt;
            flag[cnt-1]=1;
            st.erase(st.begin());
        }
        printf("%dn",x);
        for(int i=0;i<n;i++)
        {
            if(!flag[i])
            {
                num[x++]=i+1;
            }
        }
        for(int i=0;i<n;i++)
        {
            printf(i==n-1?"%dn":"%d ",num[i]);
        }
    }
    return 0;
}