hdu 1372 Knight Moves（bfs）

小記：這題看懂題意就簡單了

題意：國際象棋的馬的走法，可以有八個方向走，和中國象棋馬一樣的走法。8*8地圖上，給你一個起點一個終點，問你最少到達步數

思路：bfs

代碼：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>

using namespace std;

#define mst(a,b) memset(a,b,sizeof(a))
#define REP(a,b,c) for(int a = b; a < c; ++a)
#define eps 10e-8

const int MAX_ = 10010;
const int N = 100010;
const int INF = 0x7fffffff;

bool vis[MAX_][MAX_];

int dir[8][2] = {{1,-2}, {-1,-2},{-2,-1},{-2,1},{-1,2}, {1,2},{2,1},{2,-1}};
bool flag;
int n, m;

struct point {
    int x, y;
    int step;
}end;


int bfs(point start)
{
    queue<point> q;
    start.step = 0;
    q.push(start);

    vis[start.x][start.y] = 1;

    while(!q.empty()){
        point cur = q.front(); q.pop();

        if(cur.x == end.x && cur.y == end.y){
            //printf("%d\n", cur.step);
            return cur.step;
        }

        //printf("cur:[%d, %d]\n", cur.x, cur.y);
        REP(i, 0, 8){
            point nt;
            nt.x = cur.x + dir[i][1];
            nt.y = cur.y + dir[i][0];
            nt.step = cur.step + 1;

            if((nt.x > 0 && nt.x <= 8) && (nt.y > 0 && nt.y <= 8)){
                //printf("[%d, %d]\n", nt.x, nt.y);
                if(!vis[nt.x][nt.y]){
                    vis[nt.x][nt.y] = 1;
                    q.push(nt);
                }
            }
        }
    }
    return -1;
}



int main() {
    int T;
    char s[10], t[10];
    while(~scanf("%s%s", s, t)) {

        point start;
        start.x = s[1] - '0';
        start.y = s[0] - 'a'+1;


        end.x = t[1] - '0';
        end.y = t[0] - 'a'+1;

        //printf("coor:[%d, %d] [%d, %d]\n",start.x, start.y, end.x, end.y);

        mst(vis, 0 );
        int ans = bfs(start);


        printf("To get from %s to %s takes %d knight moves.\n", s,t,ans);

    }
    return 0;
}