Algorithm Note _196: Previous test cut lattice (Java)

Directory

1 Problem Description

2 Solutions

  1 problem description Problem Description

As shown, a number of integers are filled in 3 x 3 squares.

+--*--+--+
|10* 1|52|
+--****--+
|20|30* 1|
*******--+
| 1| 2| 3|
+--+--+--+

We cut along the star Line in the diagram and get two parts, each with a number of 60.

The requirement of the subject is to ask you to determine whether the integer in the given m x n lattice can be divided into two parts, making the numbers and the two regions equal.

If there are multiple answers, output the minimum number of squares contained in the area containing the upper-left lattice.

If it cannot be split, the output is 0.

Input Format

The program reads in two integer m n with a space partition (M,N<10).

Represents the width and height of a table.

Next is n rows, each m positive integer, separated by a space. Each integer is not greater than 10000.

output Formatoutputs an integer representing the smallest number of squares that may be included in the upper-left corner of the partition in all solutions. Sample Input 13 3
1
1
1 2 3Sample Output 13Sample Input 24 3
1 1 1 1
1 2
1 1 1Sample Output 2Ten

2 Solutions

The specific code is as follows:

ImportJava.util.Scanner; Public classMain { Public Static intN, M;  Public Static intSUM;  Public Static int[] Step = {{ -1,0},{1,0},{0,-1},{0,1}};  Public Static int[] map;  Public Static int[] visited;  Public Static intresult = 10000;  Public voidTestdfs (intXintYint[] v) {//whether the other part of the cut is connected after completionV[x][y] = 2;  for(inti = 0;i < 4;i++) {            intX1 = x + step[i][0]; intY1 = y + step[i][1]; if(X1 < 0 | | X1 >= N | | Y1 < 0 | | Y1 >=m)Continue; if(V[x1][y1] = = 0) Testdfs (x1, y1, v); }    }         Public voidDfsintStartX,intStarty,intCountintsum) {        if(Sum = = SUM/2) {            intx = 0, y = 0, step = 0; int[] v =New int[N][m];  for(inti = 0;i < n;i++) {                 for(intj = 0;j < m;j++) {                    if(Visited[i][j] = = 0) {x=i; Y=J; } V[i][j]=Visited[i][j];            }} testdfs (x, y, v);  for(inti = 0;i < n;i++)                 for(intj = 0;j < m;j++)                    if(V[i][j] = = 2) Step++; if(step + Count = = n *m) {if(Visited[0][0] = = 1) Result=math.min (result, count); Else{result=math.min (result, step); }            }            return; } Else if(Sum > SUM/2)            return;  for(inti = 0;i < 4;i++) {            intX1 = StartX + step[i][0]; intY1 = Starty + step[i][1]; if(X1 < 0 | | X1 >= N | | Y1 < 0 | | Y1 >=m)Continue; if(Visited[x1][y1] = = 1)                Continue; VISITED[X1][Y1]= 1; Count++; Sum+=Map[x1][y1];            DFS (x1, y1, count, sum); Sum-=Map[x1][y1]; Count--; VISITED[X1][Y1]= 0; }    }         Public Static voidMain (string[] args) {main test=NewMain (); Scanner in=NewScanner (system.in); M=In.nextint (); N=In.nextint (); Map=New int[N][m]; SUM= 0;  for(inti = 0;i < n;i++) {             for(intj = 0;j < m;j++) {Map[i][j]=In.nextint (); SUM+=Map[i][j]; }        }        if(SUM% 2 = = 0) {             for(inti = 0;i < n;i++) {                 for(intj = 0;j < m;j++) {visited=New int[N][m]; intsum =Map[i][j]; intCount = 1; VISITED[I][J]= 1;                Test.dfs (i, J, Count, sum); }            }            if(Result! = 10000) System.out.println (result); ElseSystem.out.println ("-1"); } ElseSystem.out.println ("-1"); }}

Algorithm Note _196: Previous test cut lattice (Java)