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  • hdu-5437 Alisha’s Party (优先队列)

    Alisha’s Party

    Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 3926    Accepted Submission(s): 1017


    Problem Description
    Princess Alisha invites her friends to come to her birthday party. Each of her friends will bring a gift of some value v, and all of them will come at a different time. Because the lobby is not large enough, Alisha can only let a few people in at a time. She decides to let the person whose gift has the highest value enter first.

    Each time when Alisha opens the door, she can decide to let p people enter her castle. If there are less than p people in the lobby, then all of them would enter. And after all of her friends has arrived, Alisha will open the door again and this time every friend who has not entered yet would enter.

    If there are two friends who bring gifts of the same value, then the one who comes first should enter first. Given a query n Please tell Alisha who the nth person to enter her castle is.
     
    Input
    The first line of the input gives the number of test cases, T , where 1T15.

    In each test case, the first line contains three numbers k,m and q separated by blanks. k is the number of her friends invited where 1k150,000. The door would open m times before all Alisha’s friends arrive where 0mk. Alisha will have q queries where 1q100.

    The ith of the following k lines gives a string Bi, which consists of no more than 200 English characters, and an integer vi1vi108, separated by a blank. Bi is the name of the ith person coming to Alisha’s party and Bi brings a gift of value vi.

    Each of the following m lines contains two integers t(1tk) and p(0pk) separated by a blank. The door will open right after the tth person arrives, and Alisha will let p friends enter her castle.

    The last line of each test case will contain q numbers n1,...,nq separated by a space, which means Alisha wants to know who are the n1th,...,nqth friends to enter her castle.

    Note: there will be at most two test cases containing n>10000.
     
    Output
    For each test case, output the corresponding name of Alisha’s query, separated by a space.
     
    Sample Input
    1 5 2 3 Sorey 3 Rose 3 Maltran 3 Lailah 5 Mikleo 6 1 1 4 2 1 2 3
     
    Sample Output
    Sorey Lailah Rose
    题意:自己读吧,最近好懒,不想读题也不想写题意;
    思路:优先队列把优先级高的置顶,注意最好所有人都会进去,我就是忘了这个wa到cry;
    AC代码:
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <queue>
    using namespace std;
    int k,m,q;
    const int N=1e4+3;
    struct fx
    {
        int t,p;
    };
    fx fy[15*N];
    struct node
    {
        friend bool operator< (node n1,node n2)
        {
            if(n1.va==n2.va)return n1.indx>n2.indx;
            return n1.va<n2.va;
        }
        int va,indx;
    };
    node li[15*N];
    priority_queue <node> qu;
    int cmp(fx aa,fx bb)
    {
        return aa.t<bb.t;
    }
    int ans[15*N];
    char str[15*N][400];
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int cnt=1,start=1;
            scanf("%d%d%d",&k,&m,&q);
            for(int i=1;i<=k;i++)
            {
                scanf("%s%d",str[i],&li[i].va);
                li[i].indx=i;
            }
            for(int i=0;i<m;i++)
            {
                scanf("%d%d",&fy[i].t,&fy[i].p);
            }
            sort(fy,fy+m,cmp);
            for(int i=0;i<m;i++)
            {
                for(int j=start;j<=fy[i].t;j++)
                {
                    qu.push(li[j]);
                }
                start=fy[i].t+1;
                while((fy[i].p--)&&(!qu.empty()))
                {
                    ans[cnt++]=qu.top().indx;
                    qu.pop();
                }
            }
            for(int j=start;j<=k;j++)
            {
                qu.push(li[j]);
            }
            while(!qu.empty())
            {
                ans[cnt++]=qu.top().indx;
                qu.pop();
            }
            int pos;
            for(int i=1;i<q;i++)
            {
                scanf("%d",&pos);
                printf("%s ",str[ans[pos]]);
            }
            scanf("%d",&pos);
            printf("%s
    ",str[ans[pos]]);
        }
        return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5244665.html
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