USACO 3.3 Camelot(BFS+乱搞）

做的我好郁闷啊。。。卡到死了，想了一个暴力方法，枚举骑士和国王相遇的点，再枚举所有的骑士相遇的点。这样复杂度基本上上亿了，然后卡了几天，看了看题解，枚举国王和骑士的相遇的时候只需要枚举国王周围的几个点就行了。。。然后原来写的代码，各种BUG啊。。。各种调试，各种提交，各种错误n,m看错各种变量写错，实在是无语，最后终于检查不出错误了，提交还是挂了。。。找AC代码 换函数。。。最后还是没检查出那里错了，换了跟AC代码的写法，终于过了。。。这题真心不想做第二遍了。。。

/*
    ID: cuizhe
    LANG: C++
    TASK: camelot
*/
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
using namespace std;
#define INF 0x3f3f3f3f
int o[51][51],na[31][31][31][31],n,m,num;
int a[8] = {1,1,-1,-1,2,2,-2,-2};
int b[8] = {2,-2,2,-2,1,-1,1,-1};
int r[1001],c[1001];
void bfs(int x,int y)//预处理所有的可以走的最短路
{
    int quer[5001],quec[5001];
    int str,end,i,j,k;
    memset(o,-1,sizeof(o));
    quer[1] = x,quec[1] = y;
    o[x][y] = 0;
    str = end = 1;
    while(str <= end)
    {
        j = 1;
        for(i = str; i <= end; i ++)
        {
            for(k = 0; k <= 7; k ++)
            {
                if(quer[i]+a[k]<=m&&quer[i]+a[k]>=1&&quec[i]+b[k]>=1&&quec[i]+b[k]<=n&&o[quer[i]+a[k]][quec[i]+b[k]] == -1)
                {
                    o[quer[i]+a[k]][quec[i]+b[k]] = o[quer[i]][quec[i]]+1;
                    quer[end+j] = quer[i] + a[k];
                    quec[end+j] = quec[i] + b[k];
                    j ++;
                }
            }
        }
        str = end + 1;
        end = end + j - 1;
    }
    for(i = 1; i <= m; i ++)
    {
        for(j = 1; j <= n; j ++)
        {
            if(na[x][y][i][j] > o[i][j]&&o[i][j] != -1)
                na[x][y][i][j] = o[i][j];
        }
    }
}
int Abs(int x)
{
    if(x < 0)
        return -x;
    else
        return x;
}
int Max(int a,int b)
{
    return a > b ? a:b;
}
int ac(int x,int y)//枚举国王在(x,y)和骑士相遇，本来没有写成函数的。。。
{
    int i,j,k,sum = 0,ans;
    for(k = 2; k <= num; k ++)
    {
        if(na[r[k]][c[k]][x][y] == INF)//这里注意越界。。。
        return INF;
        sum += na[r[k]][c[k]][x][y];
    }
    ans = sum + Max(Abs(r[1]-x),Abs(c[1]-y));
    for(i = 2; i <= num; i ++)
    {
        for(j = Max(1,r[1]-2); j <= m&&j <= r[1]+2; j ++)
        {
            for(k = Max(1,c[1]-2); k <= n&&k <= c[1]+2;k ++)
            {
            ans=min(ans,sum-na[r[i]][c[i]][x][y]+na[r[i]][c[i]][j][k]+na[j][k][x][y]+max(Abs(r[1]-j),Abs(c[1]-k)));
            }
        }
    }
    return ans;
}
int main()
{
    int i,j,k,u,ans;
    char p[1001][3];
    freopen("camelot.in","r",stdin);
    freopen("camelot.out","w",stdout);
    scanf("%d%d",&n,&m);//其实n,m我弄反了。。。
    i = 1;
    while(scanf("%s %d",p[i],&c[i])!=EOF)
    {
        r[i] = p[i][0]-'A'+1;
        i ++;
    }
    num = i-1;
    if(num == 0)
    {
        printf("0\n");
        return 0;
    }
    for(i = 1; i <= m; i ++)
    {
        for(j = 1; j <= n; j ++)
        {
            for(k = 1; k <= m; k ++)
            {
                for(u = 1; u <= n; u ++)
                    na[i][j][k][u] = INF;
            }
            na[i][j][i][j] = 0;
        }
    }
    for(i = 1; i <= m; i ++)
    {
        for(j = 1; j <= n; j ++)
        {
            bfs(i,j);
        }
    }
    ans = INF;
    for(i = 1; i <= m; i ++)
    {
        for(j = 1; j <= n; j ++)
        {
            ans=min(ans,ac(i,j));
        }
    }
    if(ans == INF)
        printf("0\n");
    else
        printf("%d\n",ans);
    return 0;
}