Ural 1146. Maximum Sum

1146. Maximum Sumtime limit:0.5 Second
Memory limit:64 Mbgiven A 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest Sum. The sum of a rectangle is the sum of the "the elements in" that rectangle. The problem the sub-rectangle with the largest sum are referred to as the Maximal sub-rectangle. A Sub-rectangle is any contiguous sub-array of size 1x1 or greater located within the whole array. As an example, the maximal sub-rectangle of the array:

0	−2	−7	0
9	2	−6	2
−4	1	−4	1
−1	8	0	−2

is in the Lower-left-hand corner and have the sum of 15.InputThe input consists of an NX NArray of integers. The input begins with a single positive integer NOn a line by itself indicating the size of the square and the dimensional array. This was followed by N2 integers separated by white-space (newlines and spaces). These N2 integers make to the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on th e second row, left-to-right, etc.). NMay is as large as 100. The numbers in the array would be in the range [−127, 127]. Outputthe output is the sum of the maximal sub-rectangle. Sample

input	Output
40-2-7 09 2-6 2-4 1-4 1-1 8 0-2	15

Tags:Dynamic programming()difficulty:97 Test Instructions: The most important sub-matrix analysis: very classical problem. Know the maximum number of sub-segments and practices. Then enumerate the upper and lower bounds of the matrix, according to the maximum number of sub-segments and practices.

1 /**2 Create by Yzx-stupidboy3 */4#include <cstdio>5#include <cstring>6#include <cstdlib>7#include <cmath>8#include <deque>9#include <vector>Ten#include <queue> One#include <iostream> A#include <algorithm> -#include <map> -#include <Set> the#include <ctime> -#include <iomanip> - using namespacestd; -typedefLong LongLL; +typedefDoubleDB; - #defineMIT (2147483647) + #defineINF (1000000001) A #defineMLL (1000000000000000001LL) at #defineSZ (x) ((int) (x). Size ()) - #defineCLR (x, y) memset (x, y, sizeof (x)) - #definePUF Push_front - #definePub push_back - #definePOF Pop_front - #definePOB pop_back in #defineFT first - #defineSD Second to #defineMk Make_pair +  -InlineintGetint () the { *     intRet =0; $     CharCh =' ';Panax Notoginseng     BOOLFlag =0; -      while(! (Ch >='0'&& Ch <='9')) the     { +         if(Ch = ='-') Flag ^=1; ACh =GetChar (); the     } +      while(Ch >='0'&& Ch <='9') -     { $RET = RET *Ten+ Ch-'0'; $Ch =GetChar (); -     } -     returnFlag? -Ret:ret; the } - Wuyi Const intN = the; the intN, Arr[n][n]; - intSum[n][n], p[n]; Wu  -InlinevoidInput () About { $scanf"%d", &n); -      for(inti =1; I <= N; i++) -          for(intj =1; J <= N; J + +) scanf ("%d", &arr[i][j]); - } A  +InlineintWork (int*arr) the { -     intRET = arr[1], cnt = arr[1]; $      for(inti =2; I <= N; i++) the     { the         if(CNT <0) CNT =Arr[i]; the         ElseCNT + =Arr[i]; theRET =Max (ret, CNT); -     } in     returnret; the } the  AboutInlinevoidSolve () the { the     intAns =-INF; the      for(inti =1; I <= N; i++) +          for(intj =1; J <= N; J + +) -SUM[I][J] = sum[i-1][J] +Arr[i][j]; the      for(inti =1; I <= N; i++)Bayi          for(intj = i; J <= N; J + +) the         { the              for(intK =1; K <= N; k++) -P[K] = sum[j][k]-sum[i-1][k]; -             intCNT =Work (p); the             /*if (ans < CNT) the             { the ans = cnt; the printf ("%d%d%d\n", ans, I, j); -             }*/ theAns =max (ans, CNT); the         } the 94cout << ans <<Endl; the } the  the intMain ()98 { AboutFreopen ("a.in","R", stdin); - Input ();101 Solve ();102     return 0;103}

View Code

Ural 1146. Maximum Sum