poj 1654--重新排序求交换的次数

把一个序列打乱，问从小到大排序后，需要经过几次交换；

Sorting by Swapping

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions:8948		Accepted: 4747

Description

Given a permutation of numbers from 1 to n, we can always get the sequence 1, 2, 3, ..., n by swapping pairs of numbers. For example, if the initial sequence is 2, 3, 5, 4, 1, we can sort them in the following way:

2 3 5 4 1

1 3 5 4 2

1 3 2 4 5

1 2 3 4 5

Here three swaps have been used. The problem is, given a specific permutation, how many swaps we needs to take at least.

Input

The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each case contains two lines. The first line contains the integer n (1 <= n <= 10000), and the second line gives the initial permutation.
Output

For each test case, the output will be only one integer, which is the least number of swaps needed to get the sequence 1, 2, 3, ..., n from the initial permutation.
Sample Input

2
3
1 2 3
5
2 3 5 4 1

Sample Output

0
3

代码：

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
#define sf scanf
#define pf printf
#define INF 1<<29
#define eps 1e-6
const double PI=acos(-1.0);
#define lint __int64
#define LL long long
const int tt=10000;
#define ULLint unsigned long long //2^64-1>1.8*10^19
#define clr(x) memset(x,0,sizeof(x))
#define Clr(x) memset(x,-1,sizeof(x))
int n;
int a[10004];
int b[10004];
int main(){
clr(a);
clr(a);
int cas;
sf("%d",&cas);
while(cas--){
sf("%d",&n);
for(int i=0;i<n;i++){
sf("%d",&a[i]);
b[i]=a[i];
}
sort(b,b+n);
int sum=0;
for(int i=0;i<n;i++){
if(a[i]!=b[i]){
for(int j=i+1;j<n;j++){
if(a[j]==b[i]){
swap(a[i],a[j]);
sum++;
}
}
}
}
pf("%d\n",sum);
}
return 0;
}