POJ 1915： Knight Moves

标签：poj

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. Can you beat him?
The Problem
Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another, so that you have the chance to be faster than Somurolov.
For people not familiar with chess, the possible knight moves are shown in Figure 1.

Input

Output

Sample Input

3
8
0 0
7 0
100
0 0
30 50
10
1 1
1 1

Sample Output

5
28
0

BFS。和2243差不多。。要你输出从起点到终点最短的步数

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>

using namespace std;

const int N = 305;
int n;
char color[N][N];
int vist[N][N];

int dx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int xx;
int yy;
struct node
{
    int x;
    int y;
    int num;
};
queue<node>q;
node a, b;

void bfs()
{
    vist[a.x][a.y]=1;
    while( !q.empty() )
    {
        node temp = q.front();
        if( temp.x==b.x && temp.y==b.y )
        {
            printf("%d\n", temp.num);
            return ;
        }
        q.pop();
        for(int i=0; i<8; i++)
        {
            xx = temp.x + dx[i];
            yy = temp.y + dy[i];
            if( xx<0 || xx>=n || yy<0 || yy>=n || vist[xx][yy])
                continue;
            node next;
            next.x = xx;
            next.y = yy;
            next.num=temp.num+1;
            vist[xx][yy]=1;
            q.push( next );
        }
    }
}

int main()
{
    int i,j;
    int cas;
    scanf("%d", &cas);
    while( cas-- )
    {
        scanf("%d", &n);
        scanf("%d%d%d%d", &a.x, &a.y, &b.x, &b.y);
        while( !q.empty() ) q.pop();
        memset( vist, 0, sizeof(vist) );
        node first;
        first.x = a.x;
        first.y = a.y;
        first.num=0;
        q.push( first );
        bfs();
    }

    return 0;
}

POJ 1915： Knight Moves,布布扣,bubuko.com

POJ 1915： Knight Moves

标签：poj

原文地址：http://blog.csdn.net/u013487051/article/details/37886167