骑士的移动(Knight Moves, UVa 439)

来源：互联网发布：推荐高中数学视频软件编辑：程序博客网时间：2024/05/22 07:39

本题实际上就是一次BFS，很普通，我这里利用对称性，将a-h视作行，1-8视作列，且a在上，h在下，1在左，7在右，这样符合C语言的数组习惯

#include<iostream>#include <queue>#include <string>using namespace std;const int maxn = 8+5;int dir[maxn][maxn]; //初始值设为0，表示还未访问，出发点的值设为1，以后没进一层，相应位置dir值加1，输出最后结果记得减1queue<int> qx, qy;int tox[] = {-2,-1,1,2,2,1,-1,-2};int toy[] = {1,2,2,1,-1,-2,-2,-1};void go(string s1, string s2){int x1 = s1[0] - 'a', y1 = s1[1] - '1';int x2 = s2[0] - 'a', y2 = s2[1] - '1';qx.push(x1); qy.push(y1); dir[x1][y1] = 1;while(!qx.empty())  //既然qx,qy同进同出,判断一个就好{int x = qx.front();qx.pop();int y = qy.front();qy.pop();if(x == x2 && y == y2) break;for(int i = 0; i < 8; i++){if (x + tox[i] >= 0 && x + tox[i] <= 7 && y + toy[i] >= 0 && y + toy[i] <= 7 && dir[x + tox[i]][y + toy[i]] == 0){dir[ x + tox[i] ][ y + toy[i] ] = dir[x][y] + 1;qx.push(x + tox[i]); qy.push(y + toy[i]);}}}cout<<"To get from " <<s1 << " to "<< s2<<" takes "<< dir[x2][y2] - 1<< " knight moves."<<endl;}int main(){string s1, s2;while(cin>>s1>>s2){//初始化memset(dir, 0, sizeof(dir));while (!qx.empty()) qx.pop();while(!qy.empty())qy.pop();go(s1, s2);}}

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