# HDU 4393 Throw nails（技巧性模拟）

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
`23100 1100 23 10051 12 23 34 13 4`
Sample Output
`Case #1:1 3 2Case #2:4 5 3 2 1          `
Hint
`HintThe first case:1st Second endPlayer1  100m    (BOOM!!)Player2  100mPlayer3    3m2nd Second endPlayer2  102m    Player3  103m    (BOOM!!)3rd  Second endPlayer2  104m    (BOOM!!)         `

`#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>using namespace std;#define lson k*2#define rson k*2+1#define M (t[k].l+t[k].r)/2#define ll long long#define INF 10086111int t;struct node{    int s;    int v;    int index;    friend bool operator<(node x,node y)//用于501s后的排序，不同路程路程长的在前，相同路程标号小的在前    {        if(x.s==y.s)            return x.index<y.index;        return x.s>y.s;    }}a[50005];int n;int find()//查找一个最大的index{    int i,maxx=-1,index=1008611,p;    for(i=0;i<n;i++)    {        if(maxx<a[i].s||(maxx==a[i].s&&a[i].index<index))//我wa了2发就是这个，标号小的等值也是要替换的        {            p=i;            maxx=a[i].s;            index=a[i].index;        }        a[i].s+=a[i].v;    }    if(p!=n-1)//把该节点换到最后        swap(a[p],a[n-1]);    n--;//数组大小--，每次查找就会快一些    return index;}int main(){int m,i,j;int test,p,size;node now;while(scanf("%d",&test)!=EOF)    {        for(p=1;p<=test;p++)        {            scanf("%d",&n);            for(i=0;i<n;i++)            {                scanf("%d%d",&a[i].s,&a[i].v);                a[i].index=i+1;            }            printf("Case #%d:\n",p);            printf("%d",find());            t=0;            while(n>0&&t<502)//模拟前502s的情况。。501貌似也可以            {                printf(" %d",find());                t++;            }            if(n>0)//如果n还大于0，直接sort一遍输出            {                sort(a,a+n);                for(i=0;i<n;i++)                {                    printf(" %d",a[i].index);                }            }            printf("\n");        }    }    return 0;}`

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