前言 最近板子题刷多了-- 题意 一个 \(8\times 8\) 的棋盘,问马从起点到终点的最短步数为多少. \(\sf Solution\) 要求最短路径嘛,显然 bfs 更优. 读入 这个读入处理有点麻烦-- 我们可以把表示行的字符转化为数字,即 ch-'a'+1 . 搜索 将起点入队,每次获取队首元素并相应扩展,记录步数. 搜到的第一条路径就是最短路径,直接输出 step . \(\sf Code\) #include<cstdio> #include<cstring> #…

#include<bits/stdc++.h> using namespace std; char a,c; int b,d; ][]; ]={,,,-,,-,-,-}; ]={,-,,,-,-,,-}; int main(){ while(cin>>a>>b>>c>>d){ memset(mp,-,sizeof(mp)); ,num=; mp[a-][b]=; ){ ;i<=;++i){ ;j<=;++j){ if(mp[i][j]…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] bfs模板题 [代码] /* 1.Shoud it use long long ? 2.Have you ever test several sample(at least therr) yourself? 3.Can you promise that the solution is right? At least,the main ideal 4.use the puts("") or putchar() or pr…

题意:8*8国际象棋棋盘,求马从起点到终点的最少步数. 编写时犯的错误:1.结构体内没构造.2.bfs函数里返回条件误写成起点.3.主函数里取行标时未注意书中的图. #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<sstream> #include<cctype> #include<cmath> #include…

Problem Description A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks tha…

这道题曾经写过,bfs.用队列,不多说了,上代码: #include<stdio.h> #include<stdlib.h> #include<string.h> #include<queue> using namespace std; int map[10][10]; int visit[10][10]; int dist[10][10]; int dx[8]={-2,-2,-1,-1,1,1,2,2}; int dy[8]={-1,1,-2,2,-2,2…

Knight Moves Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14125    Accepted Submission(s): 8269 Problem Description A friend of you is doing research on the Traveling Knight Problem (TKP) whe…

#include<iostream> #include<queue> #include<cstring> #include<string> #include<cstdio> using namespace std; struct Point { int x_, y_; int route; }; int dic[8][2] = {-1,2 ,1,2 ,2,1 ,2,-1 ,1,-2 ,-1,-2 ,-2,-1 ,-2,1}; int vis[10…

HDU1372:Knight Moves(BFS)   Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu   Description   A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves…

1450:[例 3]Knight Moves  题解 这道题可以用双向宽度搜索优化(总介绍在  BFS ) 给定了起始状态和结束状态,求最少步数,显然是用BFS,为了节省时间,选择双向BFS. 双向BFS,即从起点向终点搜,从终点向起点搜,扩展各自的状态,直到出现两者扩展的状态重合 优化:每次选择结点少的扩展 看一下骑士可以到达那些点呢?? 所以当然要开两个队列啦 设定: 1. dis[ i ][ a ][ b ]:队列 i ,从起点(a,b)至少多少步 2.    v[ i ][ a ][ b…