sicily 1781 Knight

Description

Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another. The possible knight moves are shown in Figure 1.

Figure 1  Possible knight moves on the board

Input

The first line contains an integer T (≤10), indicating the number of test cases.

In every case, the first line contains an integer N (≤500), indicating the size of the chess board (the entire board has size N  × N). The second and third line contain pair of integers (srcR, srcC), and (dstR, dstC), specifying the starting and ending position of the knight on the board (0 ≤ srcR, srcC, dstR, dstC ≤ N – 1).

Output

For every case, output the minimum distance on a single line. If starting point and ending point are equal, distance is 0. If the knight can’t reach the ending point, distance is -1.

Sample Input

2
1
0 0
0 0
10
0 1
8 9

Sample Output

0
6

分析：

简单的BFS查找，注意判断状态是否合法即可。

代码：

// Problem#: 1781
// Submission#: 1896080
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;

#define MAX 500

struct node{
    int x, y, z;
    node(int r, int s, int t){
        x = r;
        y = s;
        z = t;
    }
};

int n, a, b, c, d;
bool visit[MAX][MAX];
int move[8][2] = {{-2, -1}, {-1, -2}, {-2, 1}, {-1, 2}, {1, 2}, {2, 1}, {1, -2}, {2, -1}};

inline bool judge(int x, int y){
    return x >= 0 && x < n && y >= 0 && y < n;
}

int bfs(){
    queue<node> buffer;
    memset(visit, false, sizeof(visit));
    buffer.push(node(a, b, 0));
    visit[a][b] = true;
    while (!buffer.empty()) {
        node tmp = buffer.front();
        if (tmp.x == c && tmp.y == d) 
            return tmp.z;
        buffer.pop();
        for (int i = 0 ; i < 8 ; ++i) {
            int tx = tmp.x + move[i][0];
            int ty = tmp.y + move[i][1];
            int tz = tmp.z + 1;
            if (judge(tx, ty) && !visit[tx][ty]){
                buffer.push(node(tx, ty, tz));
                visit[tx][ty] = true;
            }
        }
    }
    return -1;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        cin >> n;
        cin >> a >> b >> c >> d;
        cout << bfs() << endl;
    }
    return 0;
}