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。。。

 1 /**************************************************************
 2     Problem: 1605
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:76 ms
 7     Memory:1644 kb
 8 ****************************************************************/
 9
10 #include <cstdio>
11 #include <algorithm>
12
13 #define rep(i, n) for (int (i) = 1; (i) <= (n); ++(i))
14 using namespace std;
15 const int dx[5] = {0, 1, 0, 0, -1};
16 const int dy[5] = {0, 0, 1, -1, 0};
17 const int inf = (int) 1e9;
18 const char CHAR[5] = {‘A‘, ‘W‘, ‘S‘, ‘N‘, ‘E‘};
19 int n, m, K, ans;
20 int Cx[1500], Cy[1500], Hx[1500], Hy[1500];
21 int f[40][64][64], cnt[64][64];
22 char step[40][64][64];
23
24 int main(){
25     scanf("%d%d%d", &n, &m, &K);
26     rep(i, n)
27         scanf("%d%d", Cx + i, Cy + i);
28     rep(i, m)
29         scanf("%d%d", Hx + i, Hy + i);
30     rep(i, n) rep(j, m){
31         int Dx = Cx[i] - Hx[j], Dy = Cy[i] - Hy[j];
32         if (abs(Dx) <= 30 && abs(Dy) <= 30)
33             ++cnt[31 + Dx][31 + Dy];
34     }
35     for (int k = 0; k <= K; ++k)
36         for (int i = 0; i <= 62; ++i)
37             for (int j = 0; j <= 62; ++j){
38                 f[k][i][j] = -inf;
39                 step[k][i][j] = ‘Z‘;
40             }
41     f[0][31][31] = 0;
42
43     rep(k, K) rep(i, 61)    rep(j, 61)
44         f[k][i][j] = cnt[i][j] + max(max(f[k - 1][i - 1][j], f[k - 1][i + 1][j]), max(f[k - 1][i][j - 1], f[k - 1][i][j + 1]));
45     ans = 0;
46     rep(i, 61) rep(j, 61)
47         ans = max(ans, f[K][i][j]);
48     rep(i, 61) rep(j, 61)
49         if (f[K][i][j] == ans)
50             step[K][i][j] = ‘A‘;
51     for (int k = K - 1; k >= 0; --k)
52         rep(i, 61) rep(j, 61) rep(l, 4)
53             if (f[k][i][j] + cnt[i + dx[l]][j + dy[l]] == f[k + 1][i + dx[l]][j + dy[l]] && step[k + 1][i + dx[l]][j + dy[l]] < ‘Z‘)
54                 step[k][i][j] = CHAR[l];
55
56     printf("%d\n", ans);
57     int i = 31, j = 31;
58     for (int k = 0; k < K; ++k){
59         printf("%c", step[k][i][j]);
60         if (step[k][i][j] == ‘E‘) --i; else
61         if (step[k][i][j] == ‘W‘) ++i; else
62         if (step[k][i][j] == ‘S‘) ++j; else
63         if (step[k][i][j] == ‘N‘) --j;
64     }
65     printf("\n");
66     return 0;
67 }

比某些斜率优化还长。。。