UVa 439 - Knight Moves 搜索专题
2012-07-06 01:01
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439 - Knight Moves | 13381 | 56.27% | 5157 | 93.21% |
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=105&page=show_problem&problem=380
题目类型: 搜索
样例输入:
e2 e4 a1 b2 b2 c3 a1 h8 a1 h7 h8 a1 b1 c3 f6 f6
样例输出:
To get from e2 to e4 takes 2 knight moves. To get from a1 to b2 takes 4 knight moves. To get from b2 to c3 takes 2 knight moves. To get from a1 to h8 takes 6 knight moves. To get from a1 to h7 takes 5 knight moves. To get from h8 to a1 takes 6 knight moves. To get from b1 to c3 takes 1 knight moves. To get from f6 to f6 takes 0 knight moves.
分析:
这题也是一道十分经典的搜索入门题。 由于题目是指定象棋中马的开始位置,与目标位置, 要求马以最少的步数走到目标位置。 凡是求最短步数的,一般用BFS做比较好。
求步数时, 有个小技巧, 开个vis数组,初始化为0, 然后这个用来记录走的步数,而不仅仅是用来标记是否走过。具体见代码
#include<iostream> #include<cstdio> #include<cstring> using namespace std; char start[3], end[3]; int dir[8][2] = {{-2,1},{-1,2},{1,2},{2,1}, {2,-1},{1,-2},{-1,-2},{-2,-1}}; int vis[10][10]; struct Node{int x, y; }; Node que[1000]; void bfs(){ int front=0, rear=1; que[0].x=start[0]-'a', que[0].y=start[1]-'0'-1; vis[que[0].x][que[0].y] = 1; while(front<rear){ Node t = que[front++]; // 如果遇到满足条件的,直接输出 if(t.x==end[0]-'a' && t.y==end[1]-'0'-1){ printf("To get from %s to %s takes %d knight moves.\n", start,end,vis[t.x][t.y]-1); return ; } for(int i=0; i<8; ++i){ int dx=t.x+dir[i][0], dy=t.y+dir[i][1]; if(dx>=0 && dx<8 && dy>=0 && dy<8 && !vis[dx][dy]){ vis[dx][dy] = vis[t.x][t.y]+1; // 记住步数 Node temp; temp.x=dx, temp.y=dy; que[rear++] = temp; } } } } int main(){ #ifdef LOCAL freopen("input.txt", "r", stdin); #endif while(~scanf("%s %s", start, end) && start[0]){ memset(vis, 0, sizeof(vis)); bfs(); } return 0; }
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