hdu 5360 Hiking

题意：邀请人聚会，被邀请人有两个限制that当前已经被邀请到的人数必须在[a,b]之间，否则不能被邀请，让你输出最多能邀请成功几个人，并输出一种邀请顺序

贪心+优先队列，注意一下各种边界条件。

Hiking

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1230    Accepted Submission(s): 642
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

484 1 3 2 2 1 0 35 3 6 4 2 1 7 683 3 2 0 5 0 3 64 5 2 7 7 6 7 682 2 3 3 3 0 0 27 4 3 6 3 2 2 585 6 5 3 3 1 2 46 7 7 6 5 4 3 5

 

Sample Output

71 7 6 5 2 4 3 884 6 3 1 2 5 8 773 6 7 1 5 2 8 401 2 3 4 5 6 7 8

 

Author

zimpha@zju

 

Source

2015 Multi-University Training Contest 6

#include <iostream>#include <stdio.h>#include <math.h>#include <stdlib.h>#include <string>#include <string.h>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <stack>using namespace std;typedef long long LL;const int INF=0x7fffffff;const int MAX_N=10000;int T,n,cur,pos;int ans[100009];struct people{    int x,y,id;    friend bool operator<(const people &a,const people &b){        return a.y>b.y;    }};people p[100009];bool cmp1(const people &a,const people &b){    return a.x<b.x;}priority_queue<people>q;int main(){    cin>>T;    while(T--){        memset(ans,0,sizeof(ans));        memset(p,0,sizeof(p));        while(!q.empty())q.pop();        scanf("%d",&n);        for(int i=1;i<=n;i++){            scanf("%d",&p[i].x);            p[i].id=i;        }        for(int i=1;i<=n;i++){            scanf("%d",&p[i].y);        }        sort(p+1,p+n+1,cmp1);        cur=0;        pos=1;        int backpos=n-1;        int flag=0;        while(1){            flag=0;            while(p[pos].x<=cur&&pos<=n){                flag=1;                q.push(p[pos]);                pos++;            }            while(!q.empty()){                flag=1;                if(q.top().y<cur){                    ans[backpos--]=q.top().id;                    q.pop();                    continue;                }                ans[cur]=q.top().id;                cur++;                q.pop();                break;            }            if(flag==0)break;        }        for(int i=0;i<n-(pos-1);i++){            ans[backpos--]=p[n-i].id;        }        printf("%d\n",cur);        for(int i=0;i<n-1;i++){            printf("%d ",ans[i]);        }        printf("%d\n",ans[n-1]);    }    return 0;}