HDU 5360——Hiking——————【贪心+优先队列】

Hiking

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 118    Accepted Submission(s): 69
Special Judge

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 n (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

 

 

题目大意：Bate想邀请sodas出去玩。每个sodai有一个要求，即当前已经邀请到的人数必须在[l,r]之间。问按照什么顺序邀请sodas能使最后邀请到的人最多。

 

解题思路：首先按照l从小到大排，如果l相同，按照r从小到大排。然后用优先队列保存所有l值小于等于当前人数kt的soda。优先队列是让r按照小的优先。每次从队列中取出soda，判断是否r值大于等于当前人数kt。如果是，则让kt加1，放入ans数组。如果不是，则放入b数组。如果队列中没有元素的r值大于等于当前人数kt，则说明出现中断，以后的所有soda都放入b数组。

 

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+20;
const int INF=0x3f3f3f3f;
struct Soda{
    int l,r,len,ord;
    bool operator < (const Soda &b)const {
        return r>b.r;
    }
}sodas[maxn];
priority_queue<Soda>PQ;
int ans[maxn],vis[maxn],b[maxn];    //b数组是没有去的soda编号，ans数组是去了
bool cmp(Soda a,Soda b){
    if(a.l!=b.l){
        return a.l<b.l;
    }else{
        return a.len<b.len;
    }
}
int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        int flag=0;
        for(int i=0;i<n;i++){
            scanf("%d",&sodas[i].l);
            if(sodas[i].l==0){
                flag=1;
            }
        }
        for(int i=0;i<n;i++){
            scanf("%d",&sodas[i].r);
            sodas[i].ord=i+1;
            sodas[i].len=sodas[i].r-sodas[i].l+1;
        }
        if(!flag){
            printf("0
",n);
            printf("1");
            for(int i=2;i<=n;i++){
                printf(" %d",i);
            }printf("
");
            continue;
        }
        sort(sodas,sodas+n,cmp);
        int sum=0,kk=0,kt=0,nn=0;
        int fg=0;Soda tmp;
        for(int i=0;i<n;i++){
            if(!fg){
                if(sodas[i].l<=kt){
                    PQ.push(sodas[i]);
                }else{
                    i--;
                    int mark=0;
                    while(!PQ.empty()){
                        tmp=PQ.top();
                        PQ.pop();
                        if(tmp.r>=kt){
                            mark=1;
                            kt++;
                            ans[kk++]=tmp.ord;
                            break;
                        }else{
                            b[nn++]=tmp.ord;
                        }
                    }
                    if(mark==0&&PQ.empty())
                        fg=1;
                }
            }else{
                b[nn++]=sodas[i].ord;
            }
        }
        while(!PQ.empty()){
            tmp=PQ.top();
            PQ.pop();
            if(tmp.r>=kt){
                kt++;
                ans[kk++]=tmp.ord;
            }else{
                b[nn++]=tmp.ord;
            }
        }
        printf("%d
",kt);
        printf("%d",ans[0]);
        ans[0]=0;
        for(int i=1;i<kk;i++){
            printf(" %d",ans[i]);
            ans[i]=0;
        }
        for(int i=0;i<nn;i++){
            printf(" %d",b[i]);
            b[i]=0;
        }printf("
");
    }
    return 0;
}