2016多校联赛4L (hdu5775) Bubble Sort

Bubble Sort

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 224    Accepted Submission(s): 147

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

233 1 231 2 3

 

Sample Output

Case #1: 1 1 2Case #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 3In second case, the array has already in increasing order. So the answer of every number is 0.
 

 

Author

FZU

思路：

典型的树状数组求和

首先你要推出我们要的结果就是  abs(min(a[i],i)-(a[i]+c[i]))   a[i]是数i的起始位置     c[i]是a[i]右边有多少个比i小的数

可以发现min(a[i],i)为i能达到的最左边的点        (a[i]+c[i]))为i能达到最右边的点   弄不懂先花点时间想一下为什么。（草稿纸走起!）

因为数据只有十万

我们可以看成区间1-100000

一开始每个区间都是0

然后我们可以从后往前插入数到树状数组中。

然后一波树状数组求和就好了。

看代码 你就懂了。

#include<map>#include<set>#include<cstdio>#include<string>#include<cstring>#include<iostream> #include<algorithm>#define lowerbit(x) x&(-x)using namespace std;const int maxn = 100005;int N,T;int a[maxn],b[maxn],P[maxn],c[maxn],sum[maxn];bool pd;int add(int pos){while(pos<=100003){sum[pos]+=1;pos+=lowerbit(pos);}}int getsum(int pos){int ans=0;while(pos){ans+=sum[pos];pos-=lowerbit(pos);}return ans;}int main(){cin>>T;for(int tcase=1;tcase<=T;tcase++){scanf("%d",&N);memset(sum,0,sizeof(sum));for(int i=1;i<=N;i++){scanf("%d",&P[i]);a[P[i]]=i;}for(int i=N;i>=1;i--){add(P[i]);c[P[i]]=getsum(P[i]-1);//cout<<c[P[i]]<<endl;}printf("Case #%d: ",tcase);for(int i=1;i<=N;i++){if(i==1){cout<<abs(min(a[i],i)-(a[i]+c[i]));}else cout<<" "<<abs(min(a[i],i)-(a[i]+c[i]));}cout<<endl;}}

这道题还可以用线段树来做，因为大部分能用树状数组解决地问题线段树都能解决。方法跟上面近乎一样。只是维护区间和查询方法不一样罢了。下面给代码。

#include<iostream>#include<stack>#include<cstring>#include<map>#include<string>#include<queue>#include<algorithm>#include<cstdio>#include<cmath>#include<utility>using namespace std;typedef long long LL;#define maxn 100005#define MOD 1000000007int a[maxn], b[maxn << 2],ans[maxn];void update(int l1, int r1, int l2, int r2, int now){if (l1 == r1){b[now] += 1;return;}int mid = (l1 + r1) >> 1;if (l2 <= mid)update(l1, mid, l2, r2, now << 1);if (r2>mid)update(mid + 1, r1, l2, r2, now << 1 | 1);b[now] = b[now << 1] + b[now << 1 | 1];}int query(int l1, int r1, int l2, int r2, int now){if (l2 <= l1&&r2 >= r1){return b[now];}int mid = (l1 + r1) >> 1;int a = 0, b = 0;if (l2 <= mid)a = query(l1, mid, l2, r2, now << 1);if (r2 > mid)b = query(mid + 1, r1, l2, r2, now << 1 | 1);return a + b;}int main(){int t;scanf("%d", &t);for (int tcase = 1; tcase <= t; tcase++){int n;scanf("%d", &n);memset(b, 0, sizeof(b));for (int i = 1; i <= n; i++){scanf("%d", &a[i]);}for (int i = n; i >= 1; i--){update(0, n, a[i], a[i], 1);ans[a[i]] = i + query(0, n, 0, a[i]-1, 1) - min(i, a[i]);}printf("Case #%d:", tcase);for (int i = 1; i <= n; i++){printf(" %d", ans[i]);}printf("\n");}}