POJ 2243 Knight Moves

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题目大意：

计算马在棋盘上从一个格到另一个格所走的最少步数，棋盘的行号用数字1 ~ 8表示，棋盘的列用小写字母a ~ h表示（棋盘大小为8×8），现有多个测例（以EOF结束），每个测例的输入只有一行，表示了开始位置和目标位置，形式诸如"e2 c3"，表示让你求从2行e列走到3行c列的最少步数。

题目链接

注释代码：

/*                              * Problem ID : POJ 2243 Knight Moves * Author     : Lirx.t.Una                              * Language   : C++                  * Run Time   : 79 ms                              * Run Memory : 144 KB                             */ #include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;structPoint {//表示广搜时队列中的状态charx, y;//棋盘上点的坐标charstp;//从起始位置到达该位置所走过的步数Point(void) {}Point( char xx, char yy, char sstp ) :x(xx), y(yy), stp(sstp) {}};boolvis[9][9];//表示棋盘上的点是否被访问过//马的8中走法chardx[9] = { 0, 1,  1, -1, -1, 2,  2, -2, -2 };chardy[9] = { 0, 2, -2,  2, -2, 1, -1,  1, -1 };intmain() {inti;//计数变量//start and end，临时接受起点终点坐标的变量intsx, sy;intex, ey;//临时坐标intvx, vy;Pointp;//临时点charc1, c2;//用于结束输入数据out:while ( ~scanf("\n%c%d %c%d", &c1, &sy, &c2, &ey) ) {if ( c1 == c2 && sy == ey ) {//特殊情况，起点等于终点printf("To get from %c%d to %c%d takes %d knight moves.\n",   c1, sy, c2, sy, 0);continue;}memset(vis, false, sizeof(vis));sx = c1 - 'a' + 1;//转换成好处理的数字坐标ex = c2 - 'a' + 1;queue<Point>que;vis[sx][sy] = true;//起点如队列que.push( Point( sx, sy, 0 ) );while ( !que.empty() ) {p = que.front();for ( i = 1; i < 9; i++ ) {vx = p.x + dx[i];vy = p.y + dy[i];if ( vx == ex && vy == ey ) {//找到目标printf("To get from %c%d to %c%d takes %d knight moves.\n",   c1, sy, c2, ey, p.stp + 1);goto out;}if ( vx < 1 || vx > 8 || vy < 1 || vy > 8 || vis[vx][vy] )continue;//不在棋盘中或已经被访问过que.push( Point( vx, vy, p.stp + 1 ) );vis[vx][vy] = true;}que.pop();}}return 0;}

无注释代码：

#include <iostream>#include <cstring>#include <cstdio>#include <queue>using namespace std;structPoint {charx, y;charstp;Point(void) {}Point( char xx, char yy, char sstp ) :x(xx), y(yy), stp(sstp) {}};boolvis[9][9];chardx[9] = { 0, 1,  1, -1, -1, 2,  2, -2, -2 };chardy[9] = { 0, 2, -2,  2, -2, 1, -1,  1, -1 };intmain() {inti;intsx, sy;intex, ey;intvx, vy;Pointp;charc1, c2;out:while ( ~scanf("\n%c%d %c%d", &c1, &sy, &c2, &ey) ) {if ( c1 == c2 && sy == ey ) {printf("To get from %c%d to %c%d takes %d knight moves.\n",   c1, sy, c2, sy, 0);continue;}memset(vis, false, sizeof(vis));sx = c1 - 'a' + 1;ex = c2 - 'a' + 1;queue<Point>que;vis[sx][sy] = true;que.push( Point( sx, sy, 0 ) );while ( !que.empty() ) {p = que.front();for ( i = 1; i < 9; i++ ) {vx = p.x + dx[i];vy = p.y + dy[i];if ( vx == ex && vy == ey ) {printf("To get from %c%d to %c%d takes %d knight moves.\n",   c1, sy, c2, ey, p.stp + 1);goto out;}if ( vx < 1 || vx > 8 || vy < 1 || vy > 8 || vis[vx][vy] )continue;que.push( Point( vx, vy, p.stp + 1 ) );vis[vx][vy] = true;}que.pop();}}return 0;}

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