Old Sorting
Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu
Submit Status Practice LightOJ 1166
Description
Given an array containing a permutation of 1 to n, you have to find the minimum number of swaps to sort the array in ascending order. A swap means, you can exchange any two elements of the array.
For example, let n = 4, and the array be 4 2 3 1, then you can sort it in ascending order in just 1 swaps (by swapping 4 and 1).
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case contains two lines, the first line contains an integer n (1 ≤ n ≤ 100). The next line contains nintegers separated by spaces. You may assume that the array will always contain a permutation of 1 to n.
Output
For each case, print the case number and the minimum number of swaps required to sort the array in ascending order.
Sample Input
3
4
4 2 3 1
4
4 3 2 1
4
1 2 3 4
Sample Output
Case 1: 1
Case 2: 2
Case 3: 0
题解：求转化成单调序列的最小次数；蓝桥杯那题一样。。。当初竟然没写出来。。。
有置换群的思想，对于每一个循环，只需要交换num - 1次就好了；把所有的加上就好了；
代码：
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define mem(x, y) memset(x, y, sizeof(x))
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 10010;
int vis[MAXN];
struct Node{
int pos,v;
friend bool operator < (Node a, Node b){
if(a.v != b.v){
return a.v < b.v;
}
else return a.pos < b.pos;
}
};
Node dt[MAXN];
int main(){
int N, kase = 0, T;
scanf("%d", &T);
while(T--){
scanf("%d", &N);
for(int i = 1; i <= N; i++){
scanf("%d", &dt[i].v);
dt[i].pos = i;
}
sort(dt + 1, dt + N + 1);
mem(vis, 0);
int ans = 0;
for(int i = 1; i <= N; i++){
if(!vis[i]){
int num = 0;
int j = i;
while(!vis[j]){
vis[j] = 1;
num++;
j = dt[j].pos;
}
ans += num - 1;
}
}
printf("Case %d: %d\n", ++kase, ans);
}
return 0;
}