[HDU 1372] Knight Moves

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1372

题目描述：

在一个棋盘中，给你两个点的坐标，同时，你有八种走法，当然题目没明确说，但是这种棋盘就是这种规则。

求：从第一个点都第二个点的最少步数；

注意：

１．因为坐标是字母加数字的形式，注意下输入格式

２．如果用ｓｃａｎｆ读取输入，那么别忘了加!=EOF，否则会报 Output limit error 错误

 3.　如果输入坐标的时候全部按字符格式输入的，在转成数字的时候，字母减掉 'a' ，那么数字要减掉 '1' ，这样横纵坐标的范围才能控制在　０－７

//cin字母加数字输入
#include<iostream>
#include<cstring>
#include<queue>
#include<cstdio>
using namespace std;

char c1,c2;
int sxx,exx;
int sx,sy,ex,ey;
int go[8][2] = {1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2};
bool vis[10][10];
struct node
{
    int x,y;
    int count;
};

bool check(node s)
{
    if(s.x>=0&&s.x<8&&s.y>=0&&s.y<8)
        return 1;
    else
        return 0;
}
void bfs()
{
    queue<node> Q;
    node now,nex;
    now.x = sx;now.y = sy;
    now.count = 0;
    vis[now.x][now.y] = 1;
    Q.push(now);
    while(!Q.empty())
    {
        now = Q.front();
        Q.pop();
        if(now.x==ex&&now.y==ey)
        {
            printf("To get from %c%d to %c%d takes %d knight moves.
",c1,sxx,c2,exx,now.count);
            return ;
        }
        for(int i=0;i<8;i++)
        {
            nex.x = now.x + go[i][0];
            nex.y = now.y + go[i][1];
            if(check(nex)&&!vis[nex.x][nex.y])
            {
                vis[nex.x][nex.y] = 1;
                nex.count = now.count + 1;
                Q.push(nex);
            }
        }
    }
    cout<<"0"<<endl;
}

int main()
{
    while(cin>>c1>>sxx>>c2>>exx)
    {
        sy = c1 - 'a';
        sx = sxx - 1;
        ey = c2 - 'a';
        ex = exx - 1;
        memset(vis,0,sizeof(vis));
        bfs();
    }
    return 0;
}

//scanf字符数组输入
#include<iostream>
#include<cstring>
#include<queue>
#include<cstdio>
using namespace std;

char s1[3],s2[3];
int sx,sy,ex,ey;
int go[8][2] = {1,2,2,1,2,-1,1,-2,-1,-2,-2,-1,-2,1,-1,2};
bool vis[10][10];
struct node
{
    int x,y;
    int count;
};

bool check(node s)
{
    if(s.x>=0&&s.x<8&&s.y>=0&&s.y<8)
        return 1;
    else
        return 0;
}
void bfs()
{
    queue<node> Q;
    node now,nex;
    now.x = sx;now.y = sy;
    now.count = 0;
    vis[now.x][now.y] = 1;
    Q.push(now);
    while(!Q.empty())
    {
        now = Q.front();
        Q.pop();
        if(now.x==ex&&now.y==ey)
        {
            printf("To get from %s to %s takes %d knight moves.
",s1,s2,now.count);
            return ;
        }
        for(int i=0;i<8;i++)
        {
            nex.x = now.x + go[i][0];
            nex.y = now.y + go[i][1];
            if(check(nex)&&!vis[nex.x][nex.y])
            {
                vis[nex.x][nex.y] = 1;
                nex.count = now.count + 1;
                Q.push(nex);
            }
        }
    }
    cout<<"0"<<endl;
}

int main()
{
    while(scanf("%s%s",s1,s2)!=EOF)
    {
        sy = s1[0] - 'a';
        sx = s1[1] - '1';
        ey = s2[0] - 'a';
        ex = s2[1] - '1';
        memset(vis,0,sizeof(vis));
        bfs();
        getchar();
    }
    return 0;
}