第三弾が始まる！
codechef problems 第三弹
一、Motorbike Racing
题面
It‘s time for the annual exciting Motorbike Race in Byteland.
There are N motorcyclists taking part in the competition. Johnny is watching the race. At the present moment (time 0), Johnny has taken note of the current velocity and position of each motorcyclist.
Johnny wants to know at a given point of time, which motorcyclist is in a specific place in the rank list. Please help him!
If at any given time two motorcyclists are in same position, the motorcyclist with the smaller index will be placed before the one with the larger index.
To make the problem simple, Johnny assumes that each motorcyclist is moving at a constant velocity.
Input
The first line contains a number t (about 10) which is the number of test cases. Then t test cases follow. Each test case has the following form.
The first line of the test case contains a number N (1 <= N <= 2000), the number of motorcyclists.
The i-th line in the next N lines contains two numbers, v and x, which are the velocity and the current position of the i-th motorcyclist (1 <= v, x <= 100,000).
The next line contains a number Q (1 <= Q <= 2000), the number of time queries.
Each line in the next Q lines contains two numbers, t (1 <= t <= 1,000,000,000) and k (1 <= k <= n), representing the query: "at time t, which motorcyclist is positioned k-th in the rank list?"
Output
For each test case, print Q lines, with each line containing the index of the motorcyclist for the corresponding query.
Remember to print a new line after each test case.
Example
Input:
1 4 2 100 3 50 4 60 5 1 4 1 1 50 2 60 4 100 1
Output:
1 4 1 4
Description
一列摩托车，描述为二元组，(速度,初始位移)
给出一些询问，问在某个时刻下，位移第k大的摩托的编号
Solution
暴力去修改，之后全部排序的话，O(nqlogn)，超时
部分排序的话，用nth_element可以水过，而且很快
这里的解给出的是O(nq+n*n)的
即，离线来做
考虑到车子的变化是有限的，即超车次数不超过n*n
我们对询问排序，每次对每辆车冒泡
这样就可以正确过了
codechef上目前排名第5 哦
rank 5 code
#include <stdio.h>
#include <algorithm>
#define RG register
#define N 2010
#define MAXBUF 1<<22
#define L long long
#define gec() ((S==T&&(T=(S=B)+fread(B,1,MAXBUF,stdin),S==T))?0:*S++)
#define puc() fwrite(buff,1,itrr-buff,stdout),itrr=buff;
char B[MAXBUF],*S=B,*T=B,buff[MAXBUF],*itrr=buff;int stt[110];
template<class Type>inline void Rin(RG Type &aa){
aa=0;bool bb=0;char cc;
for(cc=gec();(cc<‘0‘||cc>‘9‘)&&cc!=‘-‘;cc=gec())
;
for(cc==‘-‘?bb=1:aa=cc-‘0‘,cc=gec();‘0‘<=cc&&cc<=‘9‘;cc=gec())
aa=aa*10+cc-‘0‘;
aa=bb?-aa:aa;
}
template<class Type>inline void Cat(RG Type aa,RG char cc=‘\n‘){
RG int O=0;
if(!aa)
*itrr++=‘0‘;
else{
aa<0?aa=-aa,*itrr++=‘-‘:1;
for(;aa;stt[++O]=aa%10,aa/=10)
;
for(;O;*itrr++=‘0‘+stt[O--])
;
}
*itrr++=cc;
}
int kase,n,m,ans[N];
struct bike{
int v,x,ind; L now;
bool operator < (bike &o)const{
return now>o.now||now==o.now&&ind<o.ind; } }a[N];
struct resq{
int t,k,ind;
bool operator < (resq &o)const{
return t<o.t; } }b[N];
#define set_file(File) freopen(#File".in","r",stdin),freopen(#File".out","w",stdout)
#define close_file() fclose(stdin),fclose(stdout)
int main(){
set_file(cc mot rac);
Rin(kase);
while(kase--){
Rin(n);
for(RG int i=1;i<=n;i++)
Rin(a[i].v),Rin(a[i].x),a[i].ind=i;
Rin(m);
for(RG int i=1;i<=m;i++)
Rin(b[i].t),Rin(b[i].k),b[i].ind=i;
std::sort(b+1,b+1+m);
for(RG int i=1;i<=n;i++)
a[i].now=(L)a[i].v*b[1].t+a[i].x;
std::sort(a+1,a+1+n);
ans[b[1].ind]=a[b[1].k].ind;
for(RG int i=2;i<=m;i++){
RG int t=b[i].t,k=b[i].k;
for(RG int j=1;j<=n;j++)
a[j].now=(L)a[j].v*t+a[j].x;
for(RG int j=2;j<=n;j++){
RG int x=j;
while(x>1&&a[x]<a[x-1]){
bike tmp=a[x];
a[x]=a[x-1];
a[x-1]=tmp;
x--;
}
}
ans[b[i].ind]=a[k].ind;
}
for(RG int i=1;i<=m;i++)
Cat(ans[i]);
if(kase)
*itrr++=‘\n‘;
}
puc();
close_file();
return 0; }