Sort it
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1940 Accepted Submission(s): 1390
Problem Description
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.
Input
The input consists of a number of test cases. 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.
Output
For each case, output the minimum times need to sort it in ascending order on a single line.
Sample Input
Sample Output
Author
WhereIsHeroFrom
Source
Recommend
yifenfei
题意是给你一个序列,问你最少交换多少次可以使整个序列单调上升。
数据只有1000,直接n^2可做,但是有一种更为精妙的方法——树状数组 0Ms
我们先初始化整个序列为0,然后每输入一个数x,就在x位置将0换成1,此时求该位置的sum,就是比x小的数字数目,显然i - sum就是是这个数字要交换的次数。
#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;
int a[1010], c[1010] , n;
int lowbit(int x)
{
return x & (-x);
}
int sum(int x)
{
int sum = 0;
while(x > 0)
{
sum += c[x];
x -= lowbit(x);
}
return sum;
}
void update(int x , int num)
{
while(x <= n)
{
c[x] += num;
x += lowbit(x);
}
}
int main()
{
while(~scanf("%d" , &n))
{
memset(a , 0 , sizeof(a));
memset(c , 0 , sizeof(c));
int ans = 0;
int x;
for(int i = 1 ; i <= n ; i++)
{
scanf("%d" , &x);
update(x , 1);
ans += i - sum(x);
}
printf("%d\n" , ans);
}
return 0;
}