[HDU][4021][24 Puzzle]

题目：http://acm.hdu.edu.cn/showproblem.php?pid=4021

先确保两个表格的0都移到4*4内（而不是在八个顶点之一上），输出N的条件是：

　　1.八个顶点完全一样。

　　2.内部的4*4可以拼成。

参考：http://topic.csdn.net/u/20081019/22/719a3ab7-359d-4d23-8677-07ec0d7b2289.html

View Code

#include <stdio.h>
#include <math.h>
#include <algorithm>

using namespace std;

int comp(int a[], int b[])
{
    int v[8] = {0,1,2,7,16,21,22,23};
    for (int i=0; i<8; i++)
    {
        if (a[v[i]] != b[v[i]])
        {
            return false;
        }
    }
    return true;
}

void move_zero(int a[])
{
    int v[8] = {0,1,2,7,16,21,22,23};
    int t[8] = {3,6,3,6,17,20,17,20};
    for (int i=0; i<8; i++)
    {
        if (a[v[i]]==0)
        {
            a[v[i]] = a[t[i]];
            a[t[i]] = 0;
        }
    }
}

int main(void)
{
    //freopen("D:/a.txt", "r", stdin);
    int ncase;
    scanf("%d", &ncase);
    for (int x=0; x<ncase; x++)
    {
        int source[24], target[24];
        for (int i=0; i<24; i++)
        {
            scanf("%d", &source[i]);
        }
        for (int i=0; i<24; i++)
        {
            scanf("%d", &target[i]);
        }
        move_zero(source);
        move_zero(target);
        if (!comp(source, target))
        {
            printf("Y\n");
            continue;
        }
        int vt[] = {3,4,5,6,8,9,10,11,12,13,14,15,17,18,19,20};
        int s[16], t[16];
        for (int i=0; i<16; i++)
        {
            s[i] = source[vt[i]];
            t[i] = target[vt[i]];
        }
        int inver_s = 0, inver_t = 0, z_s, z_t;
        for (int i=0; i<16; i++)
        {
            for (int j=0; j<16; j++)
            {
                if (s[i] > s[j] && i < j) inver_s++;
                if (t[i] > t[j] && i < j) inver_t++;
                if (s[i*4+j] == 0 && i < 4 && j < 4) z_s = i + j;
                if (t[i*4+j] == 0 && i < 4 && j < 4) z_t = i + j;
            }
        }
        //printf("%d %d %d %d\n", inver_s, inver_t, z_s, z_t);
        int sum = inver_s + inver_t + z_s +z_t;
        printf("%c\n", sum%2?'Y':'N');
    }
    return 0;
}