UVa 10422 - Knights in FEN

刚开始审错题以为是“上下左右”四个方向走,结果编出来以后,样例都没跑过,又仔细审了审题,才明白过来,骑士的走法与国际象棋骑士的走法一样,按'日'字走。

方法比较简单,因为数据量比较小,DFS回溯+哈希判重足矣。

代码如下:

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>

using namespace std;
const int MAXSIZE = 600005;
char knight[26] = "111110111100 110000100000";
char a[5][5], save[MAXSIZE][26];
int head[MAXSIZE], next[MAXSIZE], num, f_val;
int step_x[8] = {-2,2,2,-2,1,1,-1,-1};
int step_y[8] = {-1,-1,1,1,2,-2,2,-2};
int Hash()
{
    int sum = 0, seed = 131, i;
    for(i=0; i<25; i++)
        sum = sum * seed + save[num][i];
    return (sum & 0x7FFFFFFF) % 600003;
}
bool nextval()
{
    int h = Hash();
    int u = head[h];
    while(u)
    {
        if(!strcmp(save[num], save[u]))
            return false;
        u = next[u];
    }
    next[num] = head[h];
    head[h] = num;
    return true;
}
void Copy(int s) // 将二维数组转换为一维数组
{
    int tot = 0, i, j;
    for(i=0; i<5; i++)
        for(j=0; j<5; j++)
            save[s][tot++] = a[i][j];
}
void char_swap(int x1, int y1, int x2, int y2) // 交换两个位置的元素
{
    char c = a[x1][y1];
    a[x1][y1] = a[x2][y2];
    a[x2][y2] = c;
}
void dfs(int cur,int x, int y)
{
    if(cur>=10)
        return ;
    int flag = 0;
    for(int i=0; i<8; i++)
    {
        int xx = x+step_x[i];
        int yy = y+step_y[i];
        if(xx>=0 && xx<5 && yy>=0 && yy<5)
        {
            char_swap(xx, yy, x, y);
            Copy(num);
            if(!strcmp(save[num],knight))
            {
                if(f_val>cur)
                    f_val = cur;
                char_swap(xx, yy, x, y);
                return ;
            }
            if(nextval()) // 剪枝,如果与前面的棋谱相通则不必再搜下去
            {
                ++num;
                dfs(cur+1, xx, yy);
                --num;
            }
            char_swap(xx, yy, x, y);
        }
    }
}
int main()
{
#ifdef test
    freopen("sample.txt", "r", stdin);
#endif
    int t,x,y;
    scanf("%d", &t);
    while(t--)
    {
        num = 1;
        f_val = 20;
        memset(head, 0, sizeof(head));
        for(int i=0; i<5; i++)
        {
            getchar();
            for(int j=0; j<5; j++)
            {
                scanf("%c", &a[i][j]);
                if(a[i][j] == ' ')
                {
                    x = i;
                    y = j;
                }
            }
        }
        Copy(num);
        if(!strcmp(save[num],knight)) // 判断最开始的图是不是直接符合要求
            f_val = -1;
        else
            dfs(0, x, y);
        ++num;
        if(f_val < 20)
            printf("Solvable in %d move(s).\n", f_val+1);
        else
            printf("Unsolvable in less than 11 move(s).\n");
    }
    return 0;
}


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