nefu700 Car race game(离散化+线段树)_算法 car race game-优快云博客

本文介绍了一种算法，用于解决赛车游戏中计算最大超车次数的问题。通过将速度和位置离散化，再利用线段树进行查找，实现了高效计算。文章提供了完整的代码实现，并强调了避免使用MAP以防止时间超出限制。

Car race game

Time Limit 1000ms

Memory Limit 65536K

description

　　Bob is a game programming specialist. In his new car race game, there are some racers(n means the amount of racers (1<=n<=100000)) racers star from someplace(xi means Starting point coordinate),and they possible have different speed(V means speed).so it possibly takes place to overtake(include staring the same point ). now he want to calculate the maximal amount of overtaking.

input

　　The first line of the input contains an integer n-determining the number of racers.	Next n lines follow, each line contains two integer Xi and Vi.(xi means the ith racer's Starting point coordinate, Vi means the ith racer's speed.0<xi, vi<1000000).<="" font="">

output

　　For each data set in the input print on a separate line, on the standard output, the integer that represents the maximal amount of overtaking.

sample_input

sample_output

1
6
7

hint

source

思路：先把其速度或位置离散化，然后排序，用线段树查找之前会被超车的数目。

坚决不能用MAP啊，TLE了，郁闷的很啊。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
const int mm=1e5+9;
///map<int,int>mp;
class node
{
  public:int x,v;
}f[mm];
int p[mm],in[mm];
int n,kos;
bool cmp(node a,node b)
{
  if(a.x^b.x)return a.x>b.x;
  else return a.v<b.v;
}
int lowbit(int x)
{
  return x&(-x);
}
int sum(int end)
{
    int sum = 0;
    while(end > 0)
   {
      sum += in[end];
      end -= lowbit(end);
   }
   return sum;
 }
 //增加某个元素的大小
 void add(int pos, int num)
 {  if(pos==0)return;
    while(pos <= kos)
   {
      in[pos] += num;
      pos += lowbit(pos);
   }
 }
int blook(int l,int r,int num)
{
  int mid=(l+r)/2;
  while(l<=r)
  {
    if(p[mid]==num)return mid+1;
    else if(p[mid]>num)r=mid-1;
    else l=mid+1;
    mid=(l+r)/2;
  }
  return -1;
}
int main()
{ 
  while(~scanf("%d",&n))
  { /// mp.clear();
    for(int i=0;i<n;++i)
      {scanf("%d%d",&f[i].x,&f[i].v);
       p[i]=f[i].v;
      }
    sort(p,p+n);
    int ppp=0;
    for(int i=1;i<n;++i)
      if(p[i]!=p[ppp])p[++ppp]=p[i];
    kos=ppp+1;
    for(int i=0;i<n;++i)
      f[i].v=blook(0,ppp,f[i].v);
    sort(f,f+n,cmp);
    memset(in,0,sizeof(in));
    long long ans=0;
    for(int i=0;i<n;++i)
    {
      ans+=sum(f[i].v-1);
      add(f[i].v,1);
    }
    printf("%lld\n",ans);
  }
}