Alisha’s Party
Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1079 Accepted Submission(s): 297
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.
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.
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 2http://write.blog.csdn.net/postedit
1 2 3
Sample Outputhttp://write.blog.csdn.net/postedit
Sorey Lailah Rose
Source
Recommend
优先队列+模拟,en，应该还行，时间上应该不会超时了，但是下面问题来了。。。
真的是无力吐槽了，感觉。。。
因为排序的问题，一直错，一直错，一直错，结果拖累的小组最后这个题都没A掉。。。
#include<iostream>
#include<stdio.h>
#include<math.h>
#include <string>
#include<string.h>
#include<map>
#include<queue>
#include<set>
#include<utility>
#include<vector>
#include<algorithm>
#include<stdlib.h>
using namespace std;
#define eps 1e-8
#define pii pair<int,int>
#define INF 0x3f3f3f3f
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define ll long long int
char outname[150005][205];
int coun=0; struct People
{
char name[205];
int value;
int pre;
friend bool operator < (People n1, People n2)
{
if(n1.value != n2.value) return n1.value < n2.value;
return n1.pre > n2.pre;
}
} people[150005]; int main ()
{
//cout<<"&&&"<<endl;
int Case;
rd(Case);
while(Case--)
{
int now=0;
coun=0;
int n,innum,quenum;
rd(n);
rd2(innum,quenum);
for(int i=0; i<n; i++)
{
scanf("%s",&people[i].name);
rd(people[i].value);
people[i].pre = i;
} priority_queue <People> que;
int temp1,temp2;
<span style="background-color: rgb(153, 255, 153);"> for(int i=0; i<innum; i++)
{
rd2(temp1,temp2);
for(int i=now; i<temp1; i++)
que.push(people[i]);//查push和push_back区别 now=temp1;
temp2 = (temp2>que.size() ? que.size() : temp2);
for(int i=0; i<temp2 ; i++)
{
strcpy(outname[coun++],que.top().name);
que.pop();
}
}</span>
int quearr[150005],tempmax=-1;
for(int i=0; i<quenum; i++)
{
rd(quearr[i]);
if(tempmax<quearr[i]) tempmax=quearr[i];
} if( tempmax > coun+1 )
{
for(int i=now; i<n ; i++)
que.push(people[i]);
}
while(!que.empty())
{
strcpy(outname[coun++],que.top().name);
que.pop();
}
for(int i=0; i<quenum; i++)
{
printf("%s",outname[ quearr[i]-1 ]);
if(i<quenum-1) printf(" ");
}
printf("\n");
}
return 0 ;
}
增加一个排序就好了，但是好像还是模拟的有点问题，因为还有一个排序啊，这样就很慢了，所以应该用数组记录下来，遍历模拟。。。
#include<iostream>
#include<stdio.h>
#include<math.h>
#include <string>
#include<string.h>
#include<map>
#include<queue>
#include<set>
#include<utility>
#include<vector>
#include<algorithm>
#include<stdlib.h>
using namespace std;
#define eps 1e-8
#define pii pair<int,int>
#define INF 0x3f3f3f3f
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define ll long long int
char outname[150005][205];
int coun=0;
struct T{
int a,b;
}a[150005];
bool cmp(T x,T y){
return x.a<y.a;
}
struct People
{
char name[205];
int value;
int pre;
friend bool operator < (People n1, People n2)
{
if(n1.value != n2.value) return n1.value < n2.value;
return n1.pre > n2.pre;
}
} people[150005]; int main ()
{
//cout<<"&&&"<<endl;
int Case;
rd(Case);
while(Case--)
{
int now=0;
coun=0;
int n,innum,quenum;
rd(n);
rd2(innum,quenum);
for(int i=0; i<n; i++)
{
scanf("%s",&people[i].name);
rd(people[i].value);
people[i].pre = i;
} priority_queue <People> que;
int temp1,temp2;
for(int i=0; i<innum ;i++) rd2(a[i].a,a[i].b);
sort(a,a+innum,cmp); for(int i=0; i<innum; i++)
{
temp1=a[i].a,temp2=a[i].b;
for(int i=now; i<temp1; i++)
que.push(people[i]);//查push和push_back区别 now=temp1;
//temp2 = (temp2>que.size() ? que.size() : temp2);
for(int i=0; i<temp2 ; i++)
{
if(que.empty()) break;
strcpy(outname[coun++],que.top().name);
que.pop();
}
}
int quearr[150005],tempmax=-1;
for(int i=0; i<quenum; i++)
{
rd(quearr[i]);
// if(tempmax<quearr[i]) tempmax=quearr[i];
} //if( tempmax > coun+1 )
// {
for(int i=now; i<n ; i++)
que.push(people[i]);
// }
while(!que.empty())
{
strcpy(outname[coun++],que.top().name);
que.pop();
}
for(int i=0; i<quenum; i++)
{
printf("%s",outname[ quearr[i]-1 ]);
if(i<quenum-1) printf(" ");
}
printf("\n");
}
return 0 ;
}
。。。good code。。。
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <queue>
#include <map>
#include <vector>
#include <algorithm>
#include <conio.h>
#include <iostream>
using namespace std;
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define ll long long int
#define maxn 100005
#define mod 1000000007
#define pii pair<int,int>
#define maxn 150005
int n,m,t,q,x,y;
struct node
{
int k,v;
friend bool operator <(node a,node b)
{
if(a.v==b.v) return a.k>b.k;
return a.v<b.v;
}
};
char s[maxn][205];
int v[maxn];
int ans[maxn];
int ti[maxn],tot,qq[200];
int main()
{
rd(t);
getchar();
while(t--)
{
rd2(n,m);
getchar();
rd(q);
getchar();
for(int i=1; i<=n; i++)
{
scanf("%s%d",s[i],&v[i]);
}
memset(ti,0,sizeof(ti));
for(int i=1; i<=m; i++)
{
scanf("%d%d",&x,&y);
ti[x]=y;
}
tot=0;
int mt=0;
for(int i=1; i<=q; i++)
{
rd(qq[i]);
mt=mt>qq[i]?mt:qq[i];
}
priority_queue<node> que;
node nn;
int k;
for(int i=1; i<=n; i++)
{
nn.k=i;
nn.v=v[i];
que.push(nn);
if(i==n)
{
while(!que.empty())
{
nn=que.top();
que.pop();
ans[++tot]=nn.k;
if(tot>=mt) break;
}
break;
}
for(int j=1; j<=ti[i]; j++)
{
if(que.empty()) break;
nn=que.top();
que.pop();
ans[++tot]=nn.k;
}
if(tot>=mt) break;
}
for(int i=1; i<=q; i++)
{
printf("%s",s[ans[qq[i]]]);
if(i==q) printf("\n");
else printf(" ");
}
}
return 0;
}