HDU5437 结构体排序和优先队列

Alisha’s Party

Time Limit: 3000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3882    Accepted Submission(s): 1007

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  n−th  person to enter her castle is.

 

Input

The first line of the input gives the number of test cases,  T  , where  1≤T≤15 .

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  1≤k≤150,000 . The door would open m times before all Alisha’s friends arrive where  0≤m≤k . Alisha will have  q  queries where  1≤q≤100 .

The  i−th  of the following  k  lines gives a string  Bi , which consists of no more than  200  English characters, and an integer  vi ,  1≤vi≤108 , separated by a blank. Bi  is the name of the  i−th  person coming to Alisha’s party and Bi brings a gift of value  vi .

Each of the following  m  lines contains two integers  t(1≤t≤k)  and  p(0≤p≤k)  separated by a blank. The door will open right after the  t−th  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  n1−th,...,nq−th  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
   
   
   
   
   
   
   
   


   
   
   
   

从早上10点写到下午4点，终于对啦,哈哈，好开心.

代码和思路可能没有这个人的好，大家可以去看一下这个人的代码借鉴一下。http://acm.hdu.edu.cn/discuss/problem/post/reply.php?postid=25065&messageid=1&deep=0

题目大意：

       就是Ali要过生日，找K个朋友，礼物价值大的人，可以先进房间。礼物价值相同的，按照来的先后顺序进去房间。Ali会开门m次，每一此都会叫一定的人数进入城堡。最后由q个询问，询问第几个进来的人是谁

接下来说一下我遇到的问题和思路：

        难点一就是如何把进来的人通过价值的大小和先后顺序排序。

        难点二：如何使用优先队列。

        难点三：题目给出的数据是

   
   
   
   

    
    
    
    1
5 2 3
Sorey 3
Rose 3
Maltran  3
Lailah 5
Mikleo  6
1 1
4 2
1 2 3
   
   
   
   
   
   
   
   


   
   
   
   

      那么如果我的数据将m的两个顺序换一下，输出肯定还是要不变的，所以也要对m中的num进行排序。

   
   
   
   

    
    
    
    1
5 2 3
Sorey 3
Rose 3
Maltran  3
Lailah 5
Mikleo  6
4 2 
1 1
1 2 3
   
   
   
   
   
   
   
   


   
   
   
   

接下来给出我的AC代码：

#include<cstdio> #include<algorithm> #include<cstring> #include<queue> using namespace std; struct man{ char name[200 + 5]; int val, num; bool operator < (const man & r)const{ if(r.val == val)return num > r.num; else return val < r.val; } }; int k, m, q; man body[150000 + 10]; //int a[150000 + 10][2]; int query[100 + 10]; int regu[150000 + 10]; struct out{ int haoma, ent; }a[150000 + 10]; bool cmp(const out &a1, const out &a2){ return a1.haoma < a2.haoma; } void work(){ priority_queue<man> que; int res = 0, f = 1; //如果我输入是42 11 那就一定要先排序 sort(a + 1, a + 1 + m, cmp); for(int i = 1; i <= k; i++){ que.push(body[i]); if(body[i].num == a[f].haoma && f <= m){ for(int j = 1; j <= a[f].ent && !que.empty(); j++){ regu[++res] = que.top().num; que.pop(); } a[f].ent = 0; f++; } } while(!que.empty()){ regu[++res] = que.top().num; que.pop(); } for(int g = 1; g <= q; g++){ if(g == 1)printf("%s", body[regu[query[g]]].name); else printf(" %s", body[regu[query[g]]].name); } printf("\n"); } int main(){ int t; scanf("%d", &t); getchar(); while(t--){ memset(body, 0, sizeof(body)); memset(query, 0, sizeof(query)); memset(regu, 0, sizeof(regu)); scanf("%d%d%d", &k, &m, &q); getchar(); for(int i = 1; i <= k; i++){ scanf("%s %d", body[i].name, &body[i].val); body[i].num = i; getchar(); } for(int i = 1; i <= m; i++){ scanf("%d%d", &a[i].haoma, &a[i].ent); getchar(); } for(int i = 1; i <= q ; i++){ scanf("%d", &query[i]); getchar(); } work(); } return 0; } </man></queue></cstring></algorithm></cstdio>