poj 3270 Cow Sorting
题意:有一串数字,要将它排列成升序,每次可以交换两个数,交换一次的代价为两数之和。要求代价最小。
思路:读完题,没有思路,知道是关于polya的,但是也只会最简单的求有多少种染色方法,然后上网搜题解,再然后参考大神的题解,终于明白了,还是要继续学习polya啊,
大神的博客:http://www.cnblogs.com/mcflurry/archive/2012/06/21/2557798.html
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <iostream> #include <algorithm> #include <math.h> #include <queue> #define N 10004 #define INF 1000000 using namespace std ; int num[N] , pos[N*10] , vis[N] ; int minx , maxx , n ; int find () { int i , len , x , sum , t_min ; int ss = 0 ; //计数排序求循环节的长度 memset( vis , 0 , sizeof ( vis )) ; for ( i = 1 ; i <= n ; i++ ) { len = 0 ; x = i ; sum = 0 ; t_min = INF ; while ( !vis[x] ) { vis[x] = 1 ; len++ ; sum += num[x] ; if ( num[x] < t_min ) t_min = num[x] ; x = pos[num[x]] ; } if ( len > 0 ) { int s1 = sum + ( len - 2 ) * t_min ; int s2 = sum + t_min + ( len + 1 ) * minx ; ss += s1 < s2 ? s1 : s2 ; } } return ss ; } int main() { int i , j ; while ( scanf ( "%d" , &n ) != EOF ) { for ( i = 1 ; i <= n ; i++ ) scanf ( "%d" , &num[i] ); memset( pos , 0 , sizeof ( pos )) ; minx = INF ; maxx = -INF ; for ( i = 1 ; i <= n ; i++ ) { pos[num[i]] = 1 ; if ( num[i] > maxx ) maxx = num[i] ; if ( num[i] < minx ) minx = num[i] ; } for ( i = 1 ; i <= maxx ; i++ ) pos[i] += pos[i-1] ; printf ( "%d\n" , find () ); } return 0 ; }