nyoj322 sort 归并排序，树状数组

Sort

时间限制：1000 ms | 内存限制：65535 KB
难度：4
描述You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need.

For example, 1 2 3 5 4, we only need one operation : swap 5 and 4.输入The input consists of T number of test cases.(<0T<1000) Each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.输出For each case, output the minimum times need to sort it in ascending order on a single line.样例输入

2
3
1 2 3
4
4 3 2 1

样例输出

0
6

如果按冒泡排序这些O(n^2)肯定会超时，所以需要找一种更快的方法 --------归并排序。

归并排序是建立在归并操作上的一种有效的排序算法。该算法是采用分治法的一个非常典型的应用。将已有序的子序列合并，得到完全有序的序列；即先使每个子序列有序，再使子序列段间有序。若将两个有序表合并成一个有序表，称为2-路归并。这题还可以用树状数组来做

法一：用归并排序做

#include<stdio.h>
int a[1000001],b[1000001]; /*合并排序结果先保存到b中*/
int merge(int a[],int low,int mid,int high)
{
int i=low,j=mid+1,k=low;
int count=0;/*计数器*/
while((i<=mid)&&(j<=high))/*部分合并*/
{
if(a[i]<=a[j])
{
b[k++]=a[i++];
}
else
{
b[k++]=a[j++];
count+=(mid-i+1);
}
}
while(i<=mid)/*转储剩余部分*/
{
b[k++]=a[i++];
}
while(j<=high)
{
b[k++]=a[j++];

}
for(i=low;i<=high;++i)/*把b中的值复制给a*/
{
a[i]=b[i];
}
return count;
}
int sort(int a[],int low,int high)
{
int x,y,z;
int mid=(high+low)/2;
int i=low,j=mid+1;
if(low>=high)
{
return 0;
}
x=sort(a,low,mid);
y=sort(a,mid+1,high);
z=merge(a,low,mid,high);
return (x+y+z);
}
int main()
{
int ncases,n,i;
scanf("%d",&ncases);
while(ncases--)
{
scanf("%d",&n);
for(i=0;i<=n-1;i++)
{
scanf("%d",&a[i]);
}
printf("%d\n",sort(a,0,n-1));
}
return 0;
}

法二：用树状数组

不知道什么是树状数组的 就先看看树状数组吧。。链接：http://dongxicheng.org/structure/binary_indexed_tree/

#include<stdio.h>
#include<string.h>
int num[1004],n;
int lowbit(int x)
{
return x&(-x);
}
void add(int x)
//更新含有x的数组个数
{
while(x<=n)
{
num[x]++;
x+=lowbit(x);
}
}
int sum(int x)
//向下统计小于x的个数
{
int total=0;
while(x>0)
{
total+=num[x];
x-=lowbit(x);
}
return total;
}
int main()
{
int x,cases;
scanf("%d",&cases);
while(cases--)
{
scanf( "%d",&n);
memset( num,0,sizeof( num ));
int ss = 0;
for( int i = 0; i < n; ++i )
{
scanf( "%d",&x);
add(x);
ss += (i-sum( x - 1 ));
}
printf( "%d\n",ss );
}
return 0;
}