Codeforces Beta Round #3 A. Shortest path of the king 水题-优快云博客

本文提供了一个解决八皇后问题的最小路径方法，通过BFS算法帮助国王以最少的步数从当前位置移动到目标位置。

A. Shortest path of the king

题目连接：

http://www.codeforces.com/contest/3/problem/A

Description

The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, he wants to get from his current position s to square t in the least number of moves. Help him to do this.

In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).

Input

The first line contains the chessboard coordinates of square s, the second line — of square t.

Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1 to 8.

Output

In the first line print n — minimum number of the king's moves. Then in n lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.

L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.

Sample Input

a8
h1

Sample Output

7
RD
RD
RD
RD
RD
RD
RD

Hint

题意

给你一个8*8的棋盘

给你一个起始点和终点，让你输出一个最短的路径从起点到终点。

这个点可以朝着8个方向走。

题解：

显然优先考虑斜着走，然后不停朝着终点靠就行了。

无脑bfs也是兹瓷的，反正数据范围这么小。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2005;
string s,s1;
string ans[maxn];
int tot = 0;
int main()
{
    cin>>s;
    cin>>s1;
    while(s[0]<s1[0]&&s[1]<s1[1])s[0]++,s[1]++,ans[tot++]="RU";
    while(s[0]<s1[0]&&s[1]>s1[1])s[0]++,s[1]--,ans[tot++]="RD";
    while(s[0]>s1[0]&&s[1]<s1[1])s[0]--,s[1]++,ans[tot++]="LU";
    while(s[0]>s1[0]&&s[1]>s1[1])s[0]--,s[1]--,ans[tot++]="LD";
    while(s[0]<s1[0])s[0]++,ans[tot++]="R";
    while(s[0]>s1[0])s[0]--,ans[tot++]="L";
    while(s[1]<s1[1])s[1]++,ans[tot++]="U";
    while(s[1]>s1[1])s[1]--,ans[tot++]="D";
    cout<<tot<<endl;
    for(int i=0;i<tot;i++)
        cout<<ans[i]<<endl;
}

转载于:https://www.cnblogs.com/qscqesze/p/5266171.html