POJ 2481 cows 树状数组

题意很简单，需要注意的一点是当两头牛的range完全相同的时候后面的牛的Strongnumber==前面牛的Strongnumber，效率不是很高，2200+ms

#include<stdio.h>

#include<algorithm>

#include<string.h>

#include <iostream>

using namespace std;

typedef struct

{

	int x;

	int y;

	int index;

}cow;

cow cows[100010];

int N;

int tree[100010];

int ans[100010];

bool cmp(cow c,cow d)

{

	if(c.y==d.y)

	{

		return c.x<d.x;

	}

	return c.y>d.y;

}

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

void update(int index)

{

	while(index<=100010)

	{

		tree[index]++;

		index+=lowbit(index);

	}

}

int getsum(int index)

{

	int sum=0;

	while(index>0)

	{

		sum+=tree[index];

		index-=lowbit(index);

	}

	return sum;

}

int main()

{

	setbuf(stdout,NULL);

	int i,start,end;

	while(scanf("%d",&N)!=EOF)

	{

		if(N==0)break;

		memset(tree,0,sizeof(tree));

        for(i=0;i<N;i++)

        {

        	scanf("%d%d",&start,&end);

        	cows[i].index=i;

        	cows[i].x=++start;

        	cows[i].y=end;

        }



        sort(cows,cows+N,cmp);



        for(i=0;i<N;i++)

        {

        	if(cows[i].x==cows[i-1].x&&cows[i].y==cows[i-1].y)

        	{

        		ans[cows[i].index]=ans[cows[i-1].index];



        	}

        	else

        	{

        		ans[cows[i].index]=getsum(cows[i].x);

        	}



        	update(cows[i].x);

        }

        for(i=0;i<N-1;i++)

        {

        	printf("%d ",ans[i]);

        }

        printf("%d\n",ans[N-1]);

	}

	return 0;

}