树状数组

树状数组是普通数组的升级版本，举个例子在求和时候复杂度为log(n)但是普通的是n（不知道是不是复杂度还不怎么会），可以减少花费的时间，有时候也可以用来求逆序数（归并也可以 但是还没看过）。

树状数组最经典的图是这样的：

图中可以看出C数组就是树状数组，可以看出其中的规律，定义一个i，如果i是奇数那么树状数组就是本身，如果i是偶数那么就要看i%2了，如果为0的话那么就是1~i，否则就是

i和i-1。

int getbit(int i)

{

  return i&(-i);

}

这个函数可以很好的实现这个功能，如果不懂什么意思给个链接看看就好   http://zhidao.baidu.com/link?url=_LZztMQn74VM8tWCGQj4jTSQRTJfcIXmBw9APRC1OHhVJGqLIALOC1cndAiCUxH4zkBpI4dUCz_IhGauxxYa1_

这个只是第一步然后可以实现很多功能下面就举两个例子：

求和

int getsum(int i)

{

  int ans=0;

  while(i>0)

   {

      ans+=c[i];

       i-=getbit(i);

   }

     return ans;

}

区间修改 让a[2]-1：

void  update(int p,int i)//p输入2

{

     while(i<=n)

     {

        c[i]+=p;

         i+=getbit(i);

     ｝

}

POJ3067就是可以用树状数组求逆序数来做

Description

Japan plans to welcome the ACM ICPC World Finals and a lot of roads must be built for the venue. Japan is tall island with N cities on the East coast and M cities on the West coast (M <= 1000, N <= 1000). K superhighways will be build. Cities on each coast are numbered 1, 2, ... from North to South. Each superhighway is straight line and connects city on the East coast with city of the West coast. The funding for the construction is guaranteed by ACM. A major portion of the sum is determined by the number of crossings between superhighways. At most two superhighways cross at one location. Write a program that calculates the number of the crossings between superhighways.

Input

The input file starts with T - the number of test cases. Each test case starts with three numbers – N, M, K. Each of the next K lines contains two numbers – the numbers of cities connected by the superhighway. The first one is the number of the city on the East coast and second one is the number of the city of the West coast.

Output

For each test case write one line on the standard output:
Test case (case number): (number of crossings)

Sample Input

1
3 4 4
1 4
2 3
3 2
3 1

Sample Output

Test case 1: 5

很明显了 就是让你求逆序数，但是这道题目的数据很大需要64位来定义SUM，

AC代码

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
struct q
{
	int l,r;
}c[1001*1001];
int n,m,k;
long long w[1001]; 
int lowbit(int a)
{
	return a&(-a);
}
int cmp(struct q a,struct q b)
{
	if(a.l!=b.l)return a.l<b.l;
	else return a.r<b.r; 
}
void update(int p,int c)
{
	
	while(p<=m)
	{
	w[p]+=c;
	p+=lowbit(p);
		
	}
		
	
}

long long getsum(int c)
{
	long long sum=0;
	while(c>0)
	{
		sum+=w[c];
		c-=lowbit(c);
	}
	return sum;
}

int main()
{
	int t,n,i,j=0;
	while(~scanf("%d",&t))
	{
		while(t--)
		{
			scanf("%d%d%d",&n,&m,&k);
			for(i=1;i<=k;i++)
			scanf("%d%d",&c[i].l,&c[i].r);
			memset(w,0,sizeof(w));
			sort(c+1,c+1+k,cmp);
			long long ans=0;
			for(i=1;i<=k;i++)
			{
				update(c[i].r,1);
				ans+=i-getsum(c[i].r);//减掉当前W数组的值就是比它大的数目
			}
			printf("Test case %d: %I64d\n",++j,ans);
		}
	}
	return 0;
}