树状数组 HDU5775 逆序对
Bubble Sort
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2442 Accepted Submission(s): 1197
Problem Description
P is a permutation of the integers from 1 to N(index starting from 1).
Here is the code of Bubble Sort in C++.
for(int i=1;i<=N;++i)
for(int j=N,t;j>i;—j)
if(P[j-1] > P[j])
t=P[j],P[j]=P[j-1],P[j-1]=t;
After the sort, the array is in increasing order. ?? wants to know the absolute values of difference of rightmost place and leftmost place for every number it reached.
Input
The first line of the input gives the number of test cases T; T test cases follow.
Each consists of one line with one integer N, followed by another line with a permutation of the integers from 1 to N, inclusive.
limits
T <= 20
1 <= N <= 100000
N is larger than 10000 in only one case.
Output
For each test case output “Case #x: y1 y2 … yN” (without quotes), where x is the test case number (starting from 1), and yi is the difference of rightmost place and leftmost place of number i.
Sample Input
2
3
3 1 2
3
1 2 3
Sample Output
Case #1: 1 1 2
Case #2: 0 0 0
Hint
In first case, (3, 1, 2) -> (3, 1, 2) -> (1, 3, 2) -> (1, 2, 3) the leftmost place and rightmost place of 1 is 1 and 2, 2 is 2 and 3, 3 is 1 and 3 In second case, the array has already in increasing order. So the answer of every number is 0.
题目大意：
有点儿难理解，冒泡排序，每次把从后到当前位遍历，相当于把当前位之后最小的每次都放到当前位，非常有趣。要求，每个数能到达的最右边的位置和最左边的位置之差。
分析：
分析下，每个数其实都是先往右移（最少移0位）再往左移（最少移0位）到位置：那么能到达的最右的位置，就是该数应在的位置，加上该数当前位后面比该数小的个数；而能到达的最左的位置，就是当前位和应到位中更小的那一个。
重点理解加粗部分。
也可以反过来想。
（？）
代码如下：
#include
#include
#include
#include
#define lowbit(x) x&(-x)
using namespace std;
const int maxn=100020;
int n;
int c[maxn],a[maxn];
int ans[maxn];
int sum(int x)
{
int s=0;
for(;x;x-=lowbit(x)){
s+=c[x];
}
return s;
}
void add(int x,int y)
{
for(;x<=n;x+=lowbit(x)){
c[x]+=y;
}
}
int main()
{
int t,cas=0;;
scanf("%d",&t);
while(t--){
memset(c,0,sizeof(c));
memset(ans,0,sizeof(ans));
scanf("%d",&n);
for(int i=1;i<=n;++i){
scanf("%d",&a[i]);
}
for(int i=1;i<=n;++i){
add(a[i],1);
ans[a[i]]=a[i]+(i-sum(a[i]))-min(a[i],i);
printf("Case #%d:",++cas);
for(int i=1;i<=n;++i)
printf(" %d",ans[i]);
printf("\n");
}
return 0;
}