NEFU 700 Car race game 树状数组

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

2
2 1
2 2
5
2 6
9 4
3 1
4 9
9 1
7
5 5
6 10
5 6
3 10
9 10
9 5
2 2
							

sample_output

1
6
7
							

------------

将speed离散化。

按坐标从大到小，speed从小到大排序。

用树状数组求出比速度spd[i]后面比spd[i]小的数。和即为答案。

------------

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int maxn=100000;
int n;

struct CAR{
    int x;
    long long spd;
    int y;
}a[111111];

struct LS{
    int x;
    long long y;
}lss[111111];

bool cmp1(LS a,LS b)
{
    return a.y<b.y;
}

bool cmp2(CAR a,CAR b)
{
    if (a.x==b.x)
    {
        return a.y<b.y;
    }
    return a.x>b.x;
}

int tree[111111];

int lowbit(int x)
{
    return x&(-x);
}

void update(int x,int val)
{
    for (int i=x;i<=maxn;i+=lowbit(i))
    {
        tree[i]+=val;
    }
}

int query(int x)
{
    int ret=0;
    for (int i=x;i>0;i-=lowbit(i))
    {
        ret+=tree[i];
    }
    return ret;
}

int main()
{
    while (~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        memset(lss,0,sizeof(lss));
        memset(tree,0,sizeof(tree));
        for (int i=1;i<=n;i++)
        {
            scanf("%d%I64d",&a[i].x,&a[i].spd);
            lss[i].x=i;
            lss[i].y=a[i].spd;
        }

        //ÀëÉ¢
        sort(lss+1,lss+n+1,cmp1);
        int cnt=0;
        int last=-65535;
        for (int i=1;i<=n;i++)
        {
            if (lss[i].y!=last)
            {
                last=lss[i].y;
                cnt++;
                a[lss[i].x].y=cnt;
            }
            else
            {
                a[lss[i].x].y=cnt;
            }
        }

        maxn=0;
        for (int i=1;i<=n;i++) maxn=max(maxn,a[i].y);
        maxn++;

        sort(a+1,a+n+1,cmp2);

        long long ans=0;
        for (int i=1;i<=n;i++)
        {
            ans+=query(a[i].y-1);
            update(a[i].y,1);
        }
        printf("%lld\n",ans);
    }
    return 0;
}