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  • G

    来源hde4393

    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

    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!!)

    因为只有500初始位置,先排500次,剩下的按速度,按初始位置,按id排就可以

    #include<iostream>
    #include<stdio.h>
    #include<stdlib.h>
    #include <iomanip>
    #include<cmath>
    #include<float.h> 
    #include<string.h>
    #include<algorithm>
    #define sf scanf
    #define pf printf
    #define scf(x) scanf("%d",&x)
    #define scff(x,y) scanf("%d%d",&x,&y)
    #define prf(x) printf("%d
    ",x) 
    #define mm(x,b) memset((x),(b),sizeof(x))
    #include<vector>
    #include<queue>
    #include<map>
    #define rep(i,a,n) for (int i=a;i<n;i++)
    #define per(i,a,n) for (int i=a;i>=n;i--)
    typedef long long ll;
    const ll mod=1e9+7;
    const double eps=1e-8;
    const int inf=0x3f3f3f3f;
    using namespace std;
    const double pi=acos(-1.0);
    const int N=3e2+10;
    struct peo
    {
    	int speed,pos,id,visit;
    }a[50005];
    bool cmp(peo a,peo b)
    {
    	if(a.visit!=b.visit)
    	return a.visit>b.visit;
    	if(a.speed!=b.speed)
    	return a.speed>b.speed;
    	if(a.pos!=b.pos)
    	return a.pos>b.pos;
    	return a.id<b.id;
    }
    void solve(int cas,int n)
    {
    	pf("Case #%d:
    ",cas);
    	int ans=0,Max=0;
    	rep(i,0,min(n,502))
    	{
    		Max=0;
    		rep(j,0,n)
    		{
    			if(a[j].visit==0)
    			if(a[j].pos+a[j].speed*i>Max)
    			{
    				Max=a[j].pos+a[j].speed*i;
    				ans=j;
    			}
    		}
    		if(i==0)
    		pf("%d",a[ans].id);
    		else
    		pf(" %d",a[ans].id);
    		a[ans].visit=1;
    	}
    	if(n<=502)
    	{
    		pf("
    ");return;
    	}
    	sort(a,a+n,cmp);
    	rep(i,502,n)
    		pf(" %d",a[i].id);	
    	pf("
    ");
    }
    
    int main()
    {
    	int re,cas=1;;
    	scf(re);
    	while(re--)
    	{
    		int id=1,n;
    		scf(n);
    		rep(i,0,n)
    		{
    			a[i].visit=0;
    			a[i].id=id++;
    			scff(a[i].pos,a[i].speed);
    		}
    		solve(cas++,n);
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/wzl19981116/p/9499091.html
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