Topcoder SRM 525 DIV1 300

Test instructions: Give a n*m rectangle, each square is either empty, or there is gold, each time you can choose a direction of the rectangle all gold coins (up and down) to move a box, if the move after the gold coins out of the rectangle, then the gold coin disappears. Ask the least steps to make the checkered gold exactly k
( < Span class= "Mrow" id= "mathjax-span-290" > 1 ≤ n m ≤ )

Solution: Enumerate each sub-rectangle, if the sub-rectangle contains the amount of gold is exactly k, then greedy to calculate the cost of the sub-rectangle, that is, moving up and down to calculate the cost, left and right to move the cost, two times the cost is equal to the minimum number of moves *2+ the maximum number of move times

Code

#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <string>#include <iostream>#include <vector>using namespace STD;typedef Long Longllclassdropcoins{ Public:inta[ -][ -];intGetminimum ( vector<string>BintK) {memsetA0,sizeof(a)); for(inti =0; I < b.size (); i++) {intt =0; for(intj =0; J < b[0].length (); J + +) {if(B[i][j] = =' O ') ++t; A[I][J] = t;if(I >0) A[i][j] + = a[i-1][J]; }        }intAns =1e9; for(intI1 =0; I1 < B.size (); i1++) { for(intJ1 =0; J1 < b[0].length (); j1++) { for(intI2 = I1; I2 < B.size (); i2++) { for(intJ2 = J1; J2 < b[0].length (); j2++) {inta1 = A[i2][j2];if(I1 >0) A1-= a[i1-1][J2];if(J1 >0) A1-= a[i2][j1-1];if(I1 >0&& J1 >0) A1 + = a[i1-1][j1-1];if(A1 = = k) {intT1, t2, anst =0; T1 = i1, t2 = (int) b.size ()-I2-1; Anst + =2* min (t1, T2) + max (t1, T2); T1 = J1, t2 = (int) b[0].length ()-J2-1; Anst + =2* min (t1, T2) + max (t1, T2);                        ans = min (ans, anst); }                    }                }            }        }if(ans = =1e9) ans =-1;returnAns }};

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Topcoder SRM 525 DIV1 300