HDU 2689 Sort it(树状数组求逆序数)

Sort it

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3418    Accepted Submission(s): 2478

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

31 2 34 4 3 2 1 

 

Sample Output

06

 

Author

WhereIsHeroFrom

 

Source

ZJFC 2009-3 Programming Contest

题意，用冒泡排序的方法把n个数排成按升序排列需要几步。

点击打开链接

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>using namespace std;const int N = 1001;int n;int num[N];int c[N];struct node{    int x;    int id;}q[N] ;bool cmp(node a,node b){    return a.x < b.x;}int lowbit(int x){    return x&(-x);}int getsum(int x){    int s = 0;    while(x>0){        s += c[x];        x -= lowbit(x);    }    return s;}void add(int x,int y){    while(x<=n){        c[x] += y;        x += lowbit(x);    }}int main(){    while(scanf("%d",&n)!=EOF){        memset(c,0,sizeof(c));        memset(num,0,sizeof(num));        for(int i=1;i<=n;i++){            scanf("%d",&q[i].x);            q[i].id = i;        }        sort(q+1,q+1+n,cmp);        for(int i=1;i<=n;i++){            num[q[i].id] = i;        }        int sum = 0;        for(int i=1;i<=n;i++){            add(num[i],1);            sum += getsum(n) - getsum(num[i]);        }        printf("%d\n",sum);    }    return 0;}