WOJ1018-King Kong

Have you seen the movie King Kong? If you have seen it, you must be impressed by the scene of the exciting fight between dinosaurs and
King Kong,right? Though the dinosaurs have been fight off, King Kong has been injured very heavily. Considering that dinosaurs will come back
very quickly, King Kong brings a lot of stones for fear that the dinosaurs attack again.

Now King Kong has arranged the stones at random in one line. But Different alignments of these stones would be different to King Kong. If the
alignment is not the target in his mind, he will move the stones to their proper positions. Taking the physical consumption into consideration,
King Kong could swap only two stones (whose weight is a and b weight units) at one time and for each time he will consume a+b thermal units.
In order to minimize the physical consumption, King Kong should set a plan to move these stones. But this is too complex for king Kong and
he needs your help.

输入格式

There are several test cases. For each test case, it contains:
Line 1: One integer N (1<=N<=5000) which specifies the total number of stones.
Line 2: N integers (you are ensured that the absolute value of each integer is less than) which the weight of each stone initially. These numbers
specify the initial stone alignment. There is a blank between two consecutive integers.
Line 3: N integers (you are ensured that the absolute value of each integer is less than ) which the weight of each stone finally. These numbers
specify the target alignment in King Kong?s mind. There is a blank between two consecutive integers.The input will be ended by zero.

输出格式

Output the answer of the minimum total thermal units consumed by King Kong in the stone moving process.

样例输入

3
3 2 1
1 2 3
3
1 2 3
1 2 3
4
8 1 2 4
1 2 4 8
0

样例输出

4
0
17

//首先找出序列中所有的置换，然后对于每个置换有两种方法将其调成合法。  
//1.每次用置换中最小的数去和其他数换，ans=sum+min*(num-2）。  
//2.用全局最小的元素换所有元素，最后还要把这个元素换出去，ans=sum+allmin*(num+1)。  
//对于每个置换两种方案取min值相加就可以了。  
#include <stdio.h>  
#include <stdlib.h>  
#include <string.h>  
#include <math.h>  
#include<iostream>  
using namespace std;  
int h1[65537];//存储数所在的位置  
int h2[65537];//金刚想要数所在的位置  
int a[5001];  //第一行数  
int b[5001];  //第二行数  
bool sign[5001];//标记这个数是否移动到所想位置  
int main()  
{  
    //freopen("in.txt", "r", stdin);  
    int n, i, k, tp, cost, q,min;  
    while (scanf("%d", &n) != EOF && n != 0)  
    {  
        min = 65538;  
        memset(sign, 0, sizeof(sign));  
        cost = 0;  
        for (i = 1; i <= n; i++)  
        {  
            scanf("%d", &a[i]);  
            if (a[i] < min) min = a[i];  
            cost += a[i];//每个数都至少要交换一次  
            h1[a[i]] = i;  
        }  
        for (i = 1; i <= n; i++)  
        {  
            scanf("%d", &b[i]);  
            h2[b[i]] = i;  
        }  
        for (i = 1; i <= n; i++)  
            if (!sign[i])  
            {  
                tp = 65538;//记录循环中最小的数  
                q = i;  
                k = 0;//循环个数  
                do//找到循环个数和循环中最小的数  
                {  
                    sign[q] = true;  
                    if (b[q] < tp)  
                        tp = b[q];  
                    q = h1[b[q]];  
                    k++;  
                }while (q != i);  
                cost += (k - 2) * tp < tp + (k + 1) * min ? (k - 2) * tp : tp + (k + 1) * min;//两种方式中选择小的  
            }  
        printf("%d\n", cost);  
    }  
    return 0;  
}