BFS PKU 1915

标签：bfs   简单题   

渣渣的我做算法只能做水题啊。。

又是一道BFS水题

代码虐我千百遍，我待代码如初恋。我对代码是真爱啊

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

#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <math.h>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <iostream>
#include <deque>
using namespace std;
#define MAXN 300 + 5

struct node{
    int x;
    int y;
    int step;
};
int vis[MAXN][MAXN];
int n;
int fx,fy;
int xx[8] = {1,2,2,1,-1,-2,-2,-1};
int yy[8] = {2,1,-1,-2,-2,-1,1,2};

void BFS(int a,int b){
    queue<node>q;
    int i,j;
    int dx,dy;
    node front,rear;
    front.x = a;
    front.y = b;
    front.step = 0;
    vis[a][b] = 1;
    q.push(front);
    while(!q.empty()){
        front = q.front();
        q.pop();
        if(front.x==fx && front.y==fy){
            printf("%d\n",front.step);
        }
        for(i=0;i<8;i++){
            dx = front.x + xx[i];
            dy = front.y + yy[i];
            if(dx>=0&&dx<n&&dy>=0&&dy<n){
                if(vis[dx][dy] == 0){
                    rear.x = dx;
                    rear.y = dy;
                    rear.step = front.step+1;
                    vis[dx][dy] = 1;
                    q.push(rear);
                }
            }
        }
    }
}

int main(){
    int x1,y1;
    //int y1,y2;
    int T;

    while(~scanf("%d",&T)){
        while(T--){
            memset(vis,0,sizeof(vis));
            scanf("%d",&n);
            scanf("%d%d",&x1,&y1);
            scanf("%d%d",&fx,&fy);
            BFS(x1,y1);
        }
    }

    return 0;
}

BFS PKU 1915

标签：bfs   简单题   

原文地址：http://blog.csdn.net/zcr_7/article/details/39153371