usaco Camelot

这道题目确实是挺难想的——对于我而言，如果不看别人的思路，我也是想不出的。

我现在采用的思路是：

枚举国王和骑士们相遇的地点（最多26*30），及可能会带上国王的好心骑士及他们相遇的地点（坐标）。

呵呵……这就得想明白一点，如下图所示

如果某个骑士要去接国王，那么应该这个骑士应到国王附近一格距离（图中蓝色部分）或国王当前所在坐标位置（图中红色部分）。

当然也要考虑国王自己跑到终点的情况。

下面是我代码，仅供参考：

/*ID: guo geerPROG: camelotLANG: C++*/#include<iostream>#include<cstring>#include<string>#include<cstdio>#include<fstream>#include<cmath>using namespace std;int kingMove [9][2] = {{0, 1}, {1, 1}, {1, 0}, {1, -1}, {0, -1}, {-1, -1}, {-1, 0}, {-1, 1}, {0, 0}};int knightMove [8][2] = {{1, 2}, {2, 1}, {2, -1}, {1, -2}, {-1, -2}, {-2, -1}, {-2, 1}, {-1, 2}};int dist[27][31][27][31];int knightLocation[2000][2];void bfs(int board[][31], int x0, int y0, int X, int Y){     int f, r, r1;     int q[10000][2];     q[0][0] = x0, q[0][1] = y0;     board[x0][y0] = 0;     f=r=r1=0;          int steps = 1;     while(true)     {          r ++;          while(f != r)          {               int x, y;               for(int i=0; i<8; i++)               {                    x = q[f][0]+knightMove[i][0];                    y = q[f][1]+knightMove[i][1];                    if(x>0 && x<=X && y>0 && y<=Y && board[x][y] == -1)                    {                          q[++r1][0] = x;                          q[r1][1] = y;                          board[x][y] = steps;                    }               }         f++;    }    steps ++;    if(r1 < r) return ;    r = r1;     }}int main(){    freopen("camelot.in", "r", stdin);    freopen("camelot.out", "w", stdout);        int x0, y0, X, Y;    char ch;    scanf("%d %d", &Y, &X);    scanf(" %c %d", &ch, &y0);    x0 = ch-'A'+1;        memset(dist, -1, sizeof(dist));    for(int i=1; i<=X; i++)    for(int j=1; j<=Y; j++) bfs(dist[i][j], i, j, X, Y);        int x, y;    int n = 0;    while(scanf(" %c %d", &ch, &y) != EOF)    {            x = ch-'A'+1;            knightLocation[n][0] = x;            knightLocation[n][1] = y;            n ++;    }    int res = 1000000;    for(int i=1; i<=X; i++)    {        for(int j=1; j<=Y; j++)        {                int s0 = 0;                int isAllPositive = 1;                for(int k=0; k<n; k++)                 {                                              int tx, ty;                       tx = knightLocation[k][0];                       ty = knightLocation[k][1];                       s0 += dist[i][j][tx][ty];                          if(dist[i][j][tx][ty] < 0) isAllPositive = 0;                                                                      }                                 if(isAllPositive == 0) continue;                                                                            for(int k=0; k<n; k++)                {                        int tx, ty;                        tx = knightLocation[k][0];                        ty = knightLocation[k][1];                                                for(int dir=0; dir<9; dir++)                        {                            int x1 = kingMove[dir][0]+x0;                            int y1 = kingMove[dir][1]+y0;                            if(x1 < 1 || x1 > X || y1 < 1 || y1 > Y) continue;                              if(dist[i][j][tx][ty] < 0 || dist[i][j][x1][y1] < 0 || dist[x1][y1][tx][ty] < 0) continue;                                                                                int s = s0 - dist[i][j][tx][ty] + dist[i][j][x1][y1] + dist[x1][y1][tx][ty];                             if(dir != 8) s ++;                            if(res > s) res = s;                                                   }                                                if(res > max(abs(x0-tx),abs(y0-ty))+s0) res = max(abs(x0-tx),abs(y0-ty))+s0;                }        }    }    if(n == 0) res = 0;        printf("%d\n", res);        return 0;}