Codeforces development d

Given a matrix of n, m <= 300, and then finding a rectangle with a length and width greater than 2, it takes the least time to circle the border counterclockwise from the upper left corner. The calculation of time is as follows: given 3 numbers: Tu, TP, and TD, the number in the path is increased to Tu, equal to TP, otherwise TD.

Idea: directly pre-process four arrays and then directly launch brute force attacks. N4 method. However, a small constant can be 4000 + MS.

There is also an n3logn method ..

Specifically, the lower right corner is fixed, and then the upper right corner is enumerated, and a prefix and binary search are maintained through a push ..

But the constant must be much faster than the brute force ..

Code:

 1 #include <bits/stdc++.h> 2 #define M0(x) memset(x, 0, sizeof(x)) 3 #define repf(i, a, b) for(int i = (a); i <= (b); ++i) 4 #define repd(i, a, b) for(int i = (a); i >= (b); --i) 5 using namespace std; 6 int U[303][303], D[303][303], L[303][303], R[303][303], a[303][303]; 7 int n, m; 8 int tp, tu, td, t; 9 10 inline int f(const int& a, const int& b){11      return a == b ? tp : (a > b ? tu : td); 12 }13 14 void init(){15      scanf("%d%d%d", &tp, &tu, &td);16      M0(U), M0(D), M0(L), M0(R), M0(a);17      repf(i, 1, n) repf(j, 1, m) scanf("%d", &a[i][j]);18      repf(i, 1, n) repf(j, 1, m){19           D[i][j] = D[i-1][j] + f(a[i][j], a[i-1][j]);20           R[i][j] = R[i][j-1] + f(a[i][j], a[i][j-1]);21      }22      repd(i, n, 1) repd(j, m, 1){23           U[i][j] = U[i+1][j] + f(a[i][j], a[i+1][j]);24           L[i][j] = L[i][j+1] + f(a[i][j], a[i][j+1]);25      }26 }27 28 void solve(){29      int ansd = 0x3fffffff, ans = 0;30      int lx, ly, rx, ry, d;31      int tmp;32      for (int x = 1; x <= n; ++x)33          for (int y = 1; y <= m; ++y)34              for (int x1 = x + 2; x1 <= n; ++x1)35                  for (int y1 = y + 2; y1 <= m; ++y1){36                       tmp = D[x1][y1] - D[x][y1] + U[x][y] - U[x1][y];37                       tmp += (R[x][y1] - R[x][y] + L[x1][y] - L[x1][y1]);38                       d = abs(tmp - t);39                       if (d < ansd || (d == ansd && tmp < ans)){40                            lx = x, ly = y;41                            rx = x1, ry = y1;42                            ansd = d, ans = tmp;43                       }           44                  }45 //    cout <<ans << endl; 46      printf("%d %d %d %d\n", lx, ly, rx, ry);47 }48 49 int main(){50 //    freopen("a.in", "r", stdin);51     while (scanf("%d%d%d", &n, &m, &t) != EOF){52           init();53           solve();54     }55     return 0;56 }

View code

 

Codeforces development d