HDU 5437 Alisha’s Party(优先队列模拟)
2016-08-26 21:15
459 查看
思路:题目比较好懂的模拟题,用一个优先队列即可
模拟的时候要注意最后还会再开一次门,把剩下的人全部放进来,开门的时间并不一定是递增的,要自己排个序,还有就是要注意开门的时候是2 5这种数据,就是到第二个人到了开门,然后可以放5个人进来,这样不处理的话会RE
#include<bits/stdc++.h>
using namespace std;
const int maxn = 150000+7;
struct Node
{
char name[205];
int w;
int id;
bool operator < (const Node&a)const
{
if(w==a.w)
return id>a.id;
return w<a.w;
}
}p[maxn];
struct node
{
int t;
int peo;
}pp[maxn];
bool cmp(node a,node b)
{
return a.t<b.t;
}
int ans[maxn];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(ans,0,sizeof(ans));
int n,m,q;
scanf("%d%d%d",&n,&m,&q);
for(int i = 1;i<=n;i++)
{
scanf("%s%d",p[i].name,&p[i].w);
p[i].id = i;
}
for(int i = 1;i<=m;i++)
scanf("%d%d",&pp[i].t,&pp[i].peo);
sort(pp+1,pp+1+m,cmp);
priority_queue<Node>que;
int cnt = 1;int res = 1;
for(int i = 1;i<=m;i++)
{
int tim = pp[i].t;
for(;cnt<=tim;cnt++)
que.push(p[cnt]);
int lim = pp[i].peo;
for(int j = 1;j<=lim&&!que.empty();j++)
{
Node tmp = que.top();que.pop();
ans[res++]=tmp.id;
}
}
for(;cnt<=n;cnt++)
que.push(p[cnt]);
while(!que.empty())
{
Node tmp = que.top();
que.pop();
ans[res++]=tmp.id;
}
for(int i = 1;i<q;i++)
{
int x;
scanf("%d",&x);
printf("%s ",p[ans[x]].name);
}
int x;scanf("%d",&x);
printf("%s\n",p[ans[x]].name);
}
}
Description
Princess Alisha invites her friends to come to her birthday party. Each of her friends will bring a gift of some value
,
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
people
enter her castle. If there are less than
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
Please
tell Alisha who the
person
to enter her castle is.
Input
The first line of the input gives the number of test cases,
,
where
.
In each test case, the first line contains three numbers
and
separated
by blanks.
is
the number of her friends invited where
.
The door would open m times before all Alisha’s friends arrive where
.
Alisha will have
queries
where
.
The
of
the following
lines
gives a string
,
which consists of no more than
English
characters, and an integer
,
,
separated by a blank.
is
the name of the
person
coming to Alisha’s party and Bi brings a gift of value
.
Each of the following
lines
contains two integers
and
separated
by a blank. The door will open right after the
person
arrives, and Alisha will let
friends
enter her castle.
The last line of each test case will contain
numbers
separated
by a space, which means Alisha wants to know who are the
friends
to enter her castle.
Note: there will be at most two test cases containing
.
Output
For each test case, output the corresponding name of Alisha’s query, separated by a space.
Sample Input
Sample Output
模拟的时候要注意最后还会再开一次门,把剩下的人全部放进来,开门的时间并不一定是递增的,要自己排个序,还有就是要注意开门的时候是2 5这种数据,就是到第二个人到了开门,然后可以放5个人进来,这样不处理的话会RE
#include<bits/stdc++.h>
using namespace std;
const int maxn = 150000+7;
struct Node
{
char name[205];
int w;
int id;
bool operator < (const Node&a)const
{
if(w==a.w)
return id>a.id;
return w<a.w;
}
}p[maxn];
struct node
{
int t;
int peo;
}pp[maxn];
bool cmp(node a,node b)
{
return a.t<b.t;
}
int ans[maxn];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
memset(ans,0,sizeof(ans));
int n,m,q;
scanf("%d%d%d",&n,&m,&q);
for(int i = 1;i<=n;i++)
{
scanf("%s%d",p[i].name,&p[i].w);
p[i].id = i;
}
for(int i = 1;i<=m;i++)
scanf("%d%d",&pp[i].t,&pp[i].peo);
sort(pp+1,pp+1+m,cmp);
priority_queue<Node>que;
int cnt = 1;int res = 1;
for(int i = 1;i<=m;i++)
{
int tim = pp[i].t;
for(;cnt<=tim;cnt++)
que.push(p[cnt]);
int lim = pp[i].peo;
for(int j = 1;j<=lim&&!que.empty();j++)
{
Node tmp = que.top();que.pop();
ans[res++]=tmp.id;
}
}
for(;cnt<=n;cnt++)
que.push(p[cnt]);
while(!que.empty())
{
Node tmp = que.top();
que.pop();
ans[res++]=tmp.id;
}
for(int i = 1;i<q;i++)
{
int x;
scanf("%d",&x);
printf("%s ",p[ans[x]].name);
}
int x;scanf("%d",&x);
printf("%s\n",p[ans[x]].name);
}
}
Description
Princess Alisha invites her friends to come to her birthday party. Each of her friends will bring a gift of some value
,
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
people
enter her castle. If there are less than
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
Please
tell Alisha who the
person
to enter her castle is.
Input
The first line of the input gives the number of test cases,
,
where
.
In each test case, the first line contains three numbers
and
separated
by blanks.
is
the number of her friends invited where
.
The door would open m times before all Alisha’s friends arrive where
.
Alisha will have
queries
where
.
The
of
the following
lines
gives a string
,
which consists of no more than
English
characters, and an integer
,
,
separated by a blank.
is
the name of the
person
coming to Alisha’s party and Bi brings a gift of value
.
Each of the following
lines
contains two integers
and
separated
by a blank. The door will open right after the
person
arrives, and Alisha will let
friends
enter her castle.
The last line of each test case will contain
numbers
separated
by a space, which means Alisha wants to know who are the
friends
to enter her castle.
Note: there will be at most two test cases containing
.
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
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