链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=91
结题报告:因该说10多天没有A题了,今天过了一道比较水的题目,也算是来纪念一下吧,最近在刷搜索的时候感觉bfs有些题目是比较难的,象蛇和梯子那道题整整卡了我一个星期但是现在仍然没有过,标记一下,有时间在回来看一下!
这道题注意两个地方,一个是下标的值,一个是vis数组标记!
#include<cstdio> #include<cstring> #include<queue> using namespace std; const int dir[8][2] = {-2,1,-1,2,1,2,2,1,2,-1,1,-2,-1,-2,-2,-1}; int mp[9][9]; int vis[9][9]; struct node { int x,y; int step; }; queue<node>Q; int ex,ey,sx,sy; int bfs(node p) { node now,next; Q.push(p); while(!Q.empty()) { now=Q.front(); Q.pop(); for(int i=0;i<8;i++) { int xx = now.x+dir[i][0]; int yy = now.y+dir[i][1]; if(xx<1||xx>8||yy<1||yy>8) continue; if(xx == ex && yy == ey) { return now.step + 1; } if(!vis[xx][yy]) { next.x=xx; next.y=yy; next.step = now.step+1; Q.push(next); } } } } int main( ) { char str1[5],str2[5]; while(scanf("%s%s",str1,str2)!=EOF) { while(!Q.empty()) Q.pop(); sx=str1[0]-'a'+1; sy=str1[1]-'0'; ex=str2[0]-'a'+1; ey=str2[1]-'0'; if(sx==ex&&sy==ey) { printf("To get from %s to %s takes 0 knight moves.\n",str1,str2); continue; } memset(vis,0,sizeof(vis)); vis[sx][sy]=1; node p; p.x = sx; p.y = sy; p.step = 0; printf("To get from %s to %s takes %d knight moves.\n",str1,str2,bfs(p)); } }