POJ 1915 Knight Moves Knight Moves   Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 29912   Accepted: 14058 Description Background Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one pos…

1.链接地址: http://bailian.openjudge.cn/practice/1915 http://poj.org/problem?id=1915 2.题目: 总Time Limit: 1000ms Memory Limit: 65536kB Description BackgroundMr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from o…

#include<iostream> #include<stdio.h> #define MAXN 350 #include"queue" using namespace std; bool mark[MAXN][MAXN]; struct point { int x; int y; int step; point() { step = ; } }; bool bfs(point p); point b; int num; point e; point tem;…

/**< */#include <stdio.h> #include <string.h> #include <stdlib.h> #include <queue> #include <ctype.h> #define N 310 using namespace std; ][] = {{-, -}, {-, }, {-, -}, {-, }, {, -}, {, }, {, -}, {, }}; int vis[N][N], l, ex,…

 Knight Moves Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 20913   Accepted: 9702 Description Background Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fa…

拿这个经典题目开刀...........可是双向时间优势在这题上的效果不太明显 #include <iostream> #include <algorithm> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <vector> #include <set&…

Knight Moves Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 24094   Accepted: 11364 Description Background Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fast…

例题: POJ 1915 Knight Moves 骑士遍历问题(跳马问题) 在一个m*m的棋盘上,从任意一个给定的位置(sx , sy)出发,为象棋中的马找一条路通过最少的步数到达另一位置(ex ,ey),输出最少所需要的步数. 利用bfs求解. 当马在位置(x , y)的时候其后继节点(后继选择)是什么? 对于马,有八个方向可以选择,马可以跳到如下几个位置: (x+2 , y+1) , (x+1 , y+2 ) , (x-1 , y+2) , (x-2 , y+1), (x+2 , y -1…

所谓双向广度搜索指的是搜索沿两个方向同时进行:(1)正向搜索:从初始结点向目标结点方向搜索:(2)逆向搜索:从目标结点向初始结点方向搜索:当两个方向的搜索生成同一子结点时终止此搜索过程. 广度双向搜索通常有两种方法:(1)两个方向交替扩展:(2)选择结点个数较少的那个方向先扩展.方法(2)克服了两方向结点的生成速度不平衡的状态,可明显提高效率. [例1]Knight Moves (POJ 1915) Description Background Mr Somurolov, fabulous ch…

题目链接: POJ:http://poj.org/problem? id=3488 HDU:pid=1915">http://acm.hdu.edu.cn/showproblem.php?pid=1915 Description Following the account of Jules Verne, a scrambled message written by the middle age alchemist Arne Saknussemm, and deciphered by pro…