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HDU 2689Sort it 树状数组 逆序对
Sort it
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4110 Accepted Submission(s): 2920
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.
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
3
1 2 3
4
4 3 2 1
Sample Output
0
6
Author
WhereIsHeroFrom
Source
ZJFC 2009-3 Programming Contest
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题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=2689
题意：交换相邻的两个数，使得序列是上升的。问需要交换多少次。
题意：交换相邻的两个数，使得序列是上升的。问需要交换多少次。
思路：求出每个数ai前面有多少个数比ai大，或者每个数ai后面有多少个数比ai小。第一种方法只需要交换树状数组更新和求和的函数。第二种方法就是求逆向对的数量和只需要逆向输入a，直接标准的树状数组。
代码：
求左边大于等于 a[i]的数的个 数
#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int MAXN=1100;int c[MAXN];inline int Lowbit(int x){ return x&(-x);}void update(int i,int val){ for(i; i>0; i-=Lowbit(i)) c[i]+=val;}int sum(int i){ int temp=0; for(i; i<=MAXN; i+=Lowbit(i)) temp+=c[i]; return temp;}int main(){ int i,n; int a; while(~scanf("%d",&n)) { int ans=0; memset(c,0,sizeof(c)); for(i=1; i<=n; i++) { scanf("%d",&a); ans+=sum(a); update(a,1); } cout<<ans<<endl; } return 0;}
HDU 2689Sort it 树状数组 逆序对
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