【题目链接】click here~~
【题目大意】类型于中国象棋里面“马”的走法,给你两个坐标,一个初始坐标,一个最终坐标,在保证有解的情况下最小的步数
【思路】BFS的话,直接模拟,因为棋盘比较小
(1)BFS +队列
代码:(3ms)
#include <bits/stdc++.h> using namespace std; int dir8[8][2]= {{1,2},{2,1},{-1,2},{-2,1},{1,-2},{2,-1},{-1,-2},{-2,-1}}; bool vis[10][10]; /// Memory search struct node{ int x,y,step; /// Coordinate and step } pre,last; bool ok(int dx,int dy){ if(dx>=1&&dx<=8&&dy>=1&&dy<=8) return true; return false; } int bfs(node pre,node last){ queue <node >val; while(!val.empty()) val.pop(); node top,b; vis[pre.x][pre.y]=true; val.push(pre); while(!val.empty()) { top=val.front(); val.pop(); if(top.x==last.x&&top.y==last.y) return top.step; for(int i=0; i<8; ++i){ b.x=top.x+dir8[i][0]; b.y=top.y+dir8[i][1]; if(ok(b.x,b.y)&&!vis[b.x][b.y]){ vis[b.x][b.y]=true; b.step=top.step+1; val.push(b); } } } return 0; } char c1[5],c2[5]; int res; int main(){ while(~scanf("%s %s",c1,c2)){ memset(vis,0,sizeof(vis)); pre.x=c1[0]-'a'+1;pre.y=c1[1]-'0';pre.step=0; last.x=c2[0]-'a'+1;last.y=c2[1]-'0';last.step=0; res=bfs(pre,last); printf("To get from %s to %s takes %d knight moves.\n",c1,c2,res); } return 0; } /* Sample Input e2 e4 a1 b2 b2 c3 a1 h8 a1 h7 h8 a1 b1 c3 f6 f6 Sample Output 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. */
(2)DFS:
收获:有些时候DFS没有特定的边界,需要自己设置一个合适的边界。这题推导可知,任意两点之间马踩6步之内一定能够到达,6步之内还未搜到说明绝对不是最优结果,果断退出。所以这里的res开始时最小设定为6即可,随着设的res的增大,运行时间越来越多,因为深搜可以有很多的分支,不采取较小的边界的话,可能会浪费很多时间在无用的搜索上,所以需要剪枝。
res设不同值的运行时间如下:
res = 6 39ms
........
res = 32 TLE !!!
代码:(39ms)
#include <bits/stdc++.h> using namespace std; int dir8[8][2]= {{1,2},{2,1},{-1,2},{-2,1},{1,-2},{2,-1},{-1,-2},{-2,-1}}; bool vis[10][10]; int q; inline bool ok(int dx,int dy){ if(dx>=1&&dx<=8&&dy>=1&&dy<=8) return true; return false; } void dfs(int px,int py,int lx,int ly,int deep){ if(deep>=q) return ; if(px==lx&&py==ly&&deep<q){ q=deep; return ; } for(int i=0; i<8; ++i){ int dx=px+dir8[i][0]; int dy=py+dir8[i][1]; if(ok(dx,dy)&&!vis[dx][dy]){ vis[dx][dy]=1; dfs(dx,dy,lx,ly,deep+1); vis[dx][dy]=0; } } return ; } char c1[5],c2[5]; int px,py,lx,ly; int main() { while(~scanf("%s %s",c1,c2)){ px=c1[0]-'a'+1;py=c1[1]-'0'; lx=c2[0]-'a'+1;ly=c2[1]-'0'; memset(vis,0,sizeof(vis)); q=6; vis[px][py]=1; dfs(px,py,lx,ly,0); printf("To get from %s to %s takes %d knight moves.\n",c1,c2,q); } return 0; }
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时间: 2024-10-06 07:37:00