HDU - 5775 Bubble Sort

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.InputThe 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.
OutputFor 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.

        
 求一个数列在冒泡排序中每个数字所能到达的最左和最右位置的距离，最左即为min（目标位置，当前位置），最右是后面有多少个比它小的数+当前位置。
简单模拟即可。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 
 7 using namespace std;
 8 
 9 const int maxn=100005;
10 int c[maxn],d[maxn],g[maxn],h[maxn];
11 
12 void update(int x,int v)
13 {
14     for(;x<=maxn;x+=x&(-x))
15         g[x]+=v;
16 }
17 
18 int getsum(int x)
19 {
20     int sum=0;
21     for(;x>0;x-=x&(-x))
22         sum+=g[x];
23     return sum;
24 }
25 
26 int main()
27 {
28     int n,T,s=1;
29     scanf("%d",&T);
30     while(T--)
31     {
32         scanf("%d",&n);
33         memset(c,0,sizeof(c));
34         memset(d,0,sizeof(d));
35         memset(g,0,sizeof(g));
36         for(int i=1;i<=n;i++)
37             scanf("%d",&c[i]);
38         for(int i=1;i<=n;i++)
39         {
40             d[c[i]]=min(c[i],i);
41         }
42         for(int i=n;i>=1;i--)
43         {
44             h[c[i]]=getsum(c[i])+i;
45             update(c[i],1);
46         }
47         printf("Case #%d: %d",s++,abs(d[1]-h[1]));
48         for(int i=2;i<=n;i++)
49         {
50             printf("% d",abs(d[i]-h[i]));
51         }
52         cout<<endl;
53     }
54     
55     
56     return 0;    
57 } 

 

转载于:https://www.cnblogs.com/xibeiw/p/7371395.html