hdu 5775 Bubble Sort (树状数组)

Bubble Sort

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 591    Accepted Submission(s): 359

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.

 

题意是给出一个由1~n组成的序列，求出模拟冒泡排序时，每个数能到达的最左位置和最右位置的差。

暴力找下规律，会发现一个数右移的次数是右边比这个数小的数的个数，所以右边的位置是当前位置加上后面比这个数小的数的个数。左边位置是排完序后的位置与一开始所在的位置中更小的那个。求右边比这个数小的数的个数用树状数组，类似白书例题LA 4329的做法。

 

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int C[maxn];
int n;

inline int lowbit(int x) {
    return x & (-x);
}

int sum(int x) {
    int ret = 0;
    while(x > 0) {
        ret += C[x]; x -= lowbit(x);
    }
    return ret;
}

void add(int x, int d) {
    while(x <= maxn) {
        C[x] += d; x += lowbit(x);
    }
}

int a[maxn];
int r[maxn], l[maxn];
int id[maxn];

int main() {
    int t;
    scanf("%d", &t);
    int cas = 0;
    while(t--) {
        memset(C, 0, sizeof(C));
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) 
            scanf("%d", &a[i]);

        for(int i = n; i >= 1; i--) {
            add(a[i], 1);
            r[i] = sum(a[i] - 1);
            id[a[i]] = i;
        }

        //for(int i = 1; i <= n; i++) printf("%d ", r[i]); puts("");
        vector<int> ans;
        for(int i = 1; i <= n; i++) {
            int ll = min(id[i], i);
            int rr = id[i] + r[id[i]];
            //printf("check %d %d %d\n", i, ll, rr);
            ans.push_back(rr - ll);
        }
        printf("Case #%d:", ++cas);
        for(int i = 0; i < ans.size(); i++) {
            printf(" %d", ans[i]);
        }puts("");
    }
}