HDU---5437-Alisha’s Party(优先队列)(2015 Changchun)
2015-09-18 20:32
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Alisha’s Party
Problem DescriptionPrincess 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
[align=left]Source[/align]
2015 ACM/ICPC Asia Regional Changchun Online
#include <iostream> #include <cstring> #include <vector> #include <queue> #include <cstdio> #include <cmath> #include <algorithm> using namespace std; #define cin1(x) scanf("%d",&x) #define cin2(x,y) scanf("%d%d",&x,&y) #define cin3(x,y,z) scanf("%d%d%d",&x,&y,&z) #define ccout(x) printf("%d ",x) #define SIZE 150005 int n,m,q; char name[SIZE][201]; char ans[SIZE][201]; int name_val[SIZE]; struct People { int poi; char nam[201]; int val; People(){}; People(int a,int b,char *c):poi(a),val(b) { strcpy(nam,c); } friend bool operator<(People a,People b) { if(a.val==b.val)return a.poi>b.poi; return a.val<b.val; } }; struct Move_in { int pos; int num; friend bool operator<(Move_in a,Move_in b) { return a.pos<b.pos; } } mov[SIZE]; priority_queue<People> ff; void solve() { int x,y,z; int i,j,k; x=y=z=0; for(i=0; i<m; ++i) { while(x<mov[i].pos) { People p(x,name_val[x],name[x]); ff.push(p); ++x; } y=0; while(y<mov[i].num && !ff.empty()) { People q; q=ff.top(); strcpy(ans[++z],q.nam); ff.pop(); ++y; if(y==n)break; } } while(x<n) { People p(x,name_val[x],name[x]); ff.push(p); ++x; } while(!ff.empty()) { People q; q=ff.top(); strcpy(ans[++z],q.nam); ff.pop(); } } int main() { int T,i,j,k,l,x,y; cin1(T); while(T--) { // ff.clear(); cin3(n,m,q); for(i=0; i<n; ++i) scanf("%s%d",name[i],&name_val[i]); for(i=0; i<m; ++i) cin2(mov[i].pos,mov[i].num); sort(mov,mov+m); solve(); while(--q) { cin1(x); cout<<ans[x]<<" "; } cin1(x); cout<<ans[x]<<endl; } return 0; }
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