刷题总结——Throw nails（hdu4393）

题目：

Problem Description

The annual school bicycle contest started. ZL is a student in this school. He is so boring because he can't ride a bike!! So he decided to interfere with the contest. He has got the players' information by previous contest video. A player can run F meters the first second, and then can run S meters every second.
Each player has a single straight runway. And ZL will throw a nail every second end to the farthest player's runway. After the "BOOM", this player will be eliminated. If more then one players are NO.1, he always choose the player who has the smallest ID.

Input

In the first line there is an integer T (T <= 20), indicates the number of test cases.
In each case, the first line contains one integer n (1 <= n <= 50000), which is the number of the players. 
Then n lines follow, each contains two integers Fi(0 <= Fi <= 500), Si (0 < Si <= 100) of the ith player. Fi is the way can be run in first second and Si is the speed after one second .i is the player's ID start from 1.

Hint

Huge input, scanf is recommended.
Huge output, printf is recommended.

Output

For each case, the output in the first line is "Case #c:".
c is the case number start from 1.
The second line output n number, separated by a space. The ith number is the player's ID who will be eliminated in ith second end.

Sample Input

2
3

100 1

100 2

3 100

5

1 1

2 2

3 3

4 1

3 4

Sample Output

Case #1:
1 3 2
Case #2:
4 5 3 2 1

Hint

Hint
The first case:

1st Second end
Player1 100m (BOOM!!)
Player2 100m
Player3 3m

2nd Second end
Player2 102m
Player3 103m (BOOM!!)

3rd Second end
Player2 104m (BOOM!!)

Source

2012 Multi-University Training Contest 10

 

题目：

　　这道题其实有一个很nb的做法··这里就不提了··网上都有··

　　首先想到这道题由于严格按淘汰顺序输出肯定要模拟淘汰的过程···但肯定不会是纯模拟···

　　看到SF偏小,然后由每次淘汰最大我们考虑消除FS中的某个影响··从而利用单调性快速求出最大值··

　　是不是和NOIP2016蚯蚓很像？

　　我们将S（因为S<=100）相等的车子塞进同一个有限队列中··按第一关键字F,第二关键字标号排序,然后每次模拟时比较每个队列的对首元素即可····

代码：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<cctype>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
inline int R(){
    char c;int f=;
    for(c=getchar();c<''||c>'';c=getchar());
    for(;c<=''&&c>='';c=getchar())    f=(f<<)+(f<<)+c-'';
    return f;
}
priority_queue<    pair<int,int> >    que[];
int T,n;
int main()
{
    //freopen("a.in","r",stdin);
    T=R();
    for(int a=;a<=T;a++)
    {
        printf("Case #%d:\n",a);
        n=R();int f,s,lim=;
        for(int i=;i<=n;i++)    f=R(),s=R(),lim=max(lim,s),que[s].push(make_pair(f,-i));
        for(int t=;t<n;t++){
            int ans,maxx=-,v;
            for(int i=;i<=lim;i++){
                if(que[i].empty())    continue;
                int f=que[i].top().first,id=-que[i].top().second;
                if(f+i*t>maxx||(f+i*t==maxx&&id<ans)){
                    maxx=f+i*t;ans=id,v=i;
                }
            }
            if(t!=n-) printf("%d ",ans);
            else printf("%d",ans);
            que[v].pop();
        }
        putchar('\n');
    }
    return ;
}