BZOJ1605 [Usaco2008 Open]Crisis on the Farm 牧场危机

标题好长&&我是权限狗，汪汪！

题没看懂的我以为这是一道极难滴题目。。。然后，然后我就看懂题了。

数据少给了一个条件K <= 30...(没这条件还做个鬼。。。)

f[k, i, j]表示走了k步，x方向移动i，y方向移动j的最大被拯救牛的数量，然后方程就很好写了，略之。

（其实是太烦了，不想写）

真是一道很烦的题目。。。不仅预处理很烦，转移很烦，连输出解也很烦。。。

 /**************************************************************
     Problem: 1605
     User: rausen
     Language: C++
     Result: Accepted
     Time:76 ms
     Memory:1644 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>

 #define rep(i, n) for (int (i) = 1; (i) <= (n); ++(i))
 using namespace std;
 const int dx[] = {, , , , -};
 const int dy[] = {, , , -, };
 const int inf = (int) 1e9;
 const char CHAR[] = {'A', 'W', 'S', 'N', 'E'};
 int n, m, K, ans;
 int Cx[], Cy[], Hx[], Hy[];
 int f[][][], cnt[][];
 char step[][][];

 int main(){
     scanf("%d%d%d", &n, &m, &K);
     rep(i, n)
         scanf("%d%d", Cx + i, Cy + i);
     rep(i, m)
         scanf("%d%d", Hx + i, Hy + i);
     rep(i, n) rep(j, m){
         int Dx = Cx[i] - Hx[j], Dy = Cy[i] - Hy[j];
         if (abs(Dx) <=  && abs(Dy) <= )
             ++cnt[ + Dx][ + Dy];
     }
     for (int k = ; k <= K; ++k)
         for (int i = ; i <= ; ++i)
             for (int j = ; j <= ; ++j){
                 f[k][i][j] = -inf;
                 step[k][i][j] = 'Z';
             }
     f[][][] = ;

     rep(k, K) rep(i, )    rep(j, )
         f[k][i][j] = cnt[i][j] + max(max(f[k - ][i - ][j], f[k - ][i + ][j]), max(f[k - ][i][j - ], f[k - ][i][j + ]));
     ans = ;
     rep(i, ) rep(j, )
         ans = max(ans, f[K][i][j]);
     rep(i, ) rep(j, )
         if (f[K][i][j] == ans)
             step[K][i][j] = 'A';
     for (int k = K - ; k >= ; --k)
         rep(i, ) rep(j, ) rep(l, )
             if (f[k][i][j] + cnt[i + dx[l]][j + dy[l]] == f[k + ][i + dx[l]][j + dy[l]] && step[k + ][i + dx[l]][j + dy[l]] < 'Z')
                 step[k][i][j] = CHAR[l];

     printf("%d\n", ans);
     int i = , j = ;
     for (int k = ; k < K; ++k){
         printf("%c", step[k][i][j]);
         if (step[k][i][j] == 'E') --i; else
         if (step[k][i][j] == 'W') ++i; else
         if (step[k][i][j] == 'S') ++j; else
         if (step[k][i][j] == 'N') --j;
     }
     printf("\n");
     return ;
 }

比某些斜率优化还长。。。