骑士精神（Knight）- NOIP2004模拟试题

一、骑士精神（Knight）

     在一个5×5的棋盘上有12个白色的骑士和12个黑色的骑士， 且有一个空位。在任何时候一个骑士都能按照骑士的走法（它可以走到和它横坐标相差为1，纵坐标相差为2或者横坐标相差为2，纵坐标相差为1的格子）移动到空位上。

        给定一个初始的棋盘，怎样才能经过移动变成如下目标棋盘：

                         

为了体现出骑士精神，他们必须以最少的步数完成任务。

 

输入文件：

第一行有一个正整数T(T<=10)，表示一共有N组数据。接下来有T个5×5的矩阵，0表示白色骑士，1表示黑色骑士，*表示空位。两组数据之间没有空行。

 

输出文件：

对于每组数据都输出一行。如果能在15步以内（包括15步）到达目标状态，则输出步数，否则输出－1。

 

Sample Input

2

10110

01*11

10111

01001

00000

01011

110*1

01110

01010

00100

 

Sample Output

7

-1

源代码如下：

/****************************************************************************************************** ** Copyright (C) 2011.07.01-2013.07.01 ** Author: famousDT <[e-mail]> ** Edit date: 2011-10-22******************************************************************************************************/#include <stdio.h>#include <stdlib.h>//abs,atof(string to float),atoi,atol,atoll#include <math.h>//atan,acos,asin,atan2(a,b)(a/b atan),ceil,floor,cos,exp(x)(e^x),fabs,log(for E),log10#include <vector>#include <queue>#include <map>#include <time.h>#include <set>#include <list>#include <stack> #include <string>#include <iostream>#include <assert.h>#include <string.h>//memcpy(to,from,count#include <ctype.h>//character process:isalpha,isdigit,islower,tolower,isblank,iscntrl,isprll#include <algorithm>using namespace std;//typedef long long ll;#define MY_PI acos(-1)#define MY_MAX(a, b) ((a) > (b) ? (a) : (b))#define MY_MIN(a, b) ((a) < (b) ? (a) : (b))#define MY_MALLOC(n, type) ((type *)malloc((n) * sizeof(type)))#define MY_ABS(a) (((a) >= 0) ? (a) : (-(a)))#define MY_INT_MAX 0x7fffffff#define LOW_BIT(a) ((a) & (-(a)))//last none zero value/*==========================================================*\| IDA*\*==========================================================*/#define STEP 15 + 1const char goal[][6] = {"11111","01111","00*11","00001","00000"};char s[6][6];int ans = 0;;int bound;int cx[] = {2, 2, 1, 1, -1, -1, -2, -2};int cy[] = {1, -1, 2, -2, 2, -2, 1, -1};int h()//与目标有几个骑士不同{int t, i, j;for (t = i = 0; i < 5; ++i)for (j = 0; j < 5; ++j)if (s[i][j] != goal[i][j]) ++t;return t;}int dfs(int x, int y, int dv, int pre_move){int hv = h ();if (hv + dv > bound) return hv + dv;if (hv == 0) {ans = 1;return dv;}int next_bound = 0x7fffffff;int i;for (i = 0; i < 8; ++i) {if (i + pre_move == 7) continue;int nx = x + cx[i];int ny = y + cy[i];if (nx >= 0 && nx < 5 && ny >= 0 && ny < 5) {swap(s[x][y], s[nx][ny]);int new_bound = dfs(nx, ny, dv + 1, i);if (ans) return new_bound;next_bound = MY_MIN(new_bound, next_bound);swap(s[x][y], s[nx][ny]);}}return next_bound;}void IDA_star(int sx, int sy){ans = 0;bound = h();while (!ans && bound <= STEP)bound = dfs(sx, sy, 0, -STEP);}int main(){    FILE *in,*out;    ans = 0;    in = fopen("Knight.in","rt");    out = fopen("Knight.out","wt");int cases;int i, j, sx, sy;char c;fscanf(in, "%d%c", &cases, &c);while (cases--) {for (i = 0; i < 5; ++i) fscanf(in, "%s", s[i]);for (i = 0; i < 5; ++i) {for (j = 0; j < 5; ++j) {if (s[i][j] == '*') {sx = i;sy = j;break;}}if (j < 5) break;}IDA_star(sx, sy);if (ans) fprintf(out, "%d\n", bound);else fprintf(out, "-1\n");}system("pause");fclose(in);    fclose(out);    return 0;}