Bzoj1605 [usaco 2008 Open] crisis on the farm crisis

Long Title & I am a permission dog, Wang!

I thought this was a very difficult question... Then, I understood the question.

The data is less than a condition k <= 30... (no such condition is left blank ...)

F [K, I, j] indicates that K steps are taken, and the X direction moves I, and the Y direction moves J to the maximum number of cattle to be saved. Then the equation is well written, just a little bit.

(In fact, it is too annoying to write)

A really annoying question... Not only is preprocessing annoying, transfer annoying, but even output solutions annoying...

 

 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; else61         if (step[k][i][j] == ‘W‘) ++i; else62         if (step[k][i][j] == ‘S‘) ++j; else63         if (step[k][i][j] == ‘N‘) --j;64     }65     printf("\n");66     return 0;67 }

View code

Longer than some slope optimization...

Bzoj1605 [usaco 2008 Open] crisis on the farm crisis