Blue Bridge Cup-cut lattice (simple Dfs)

Previous questions cut lattice time limit: 1.0s memory limit: 256.0MBProblem 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.

The output format outputs 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
10 1 52
20 30 1
1 2 3 Sample Output 13 sample input 24 3
1 1 1 1
1 30 80 2
1 1 1 100 Sample output 210

Do not know why re once.

AC Code:

#include <cstdio> #include <cstring> #include <algorithm> #define INF 0x3f3f3f3fusing namespace std; int N, m;int a[15][15];int vis[15][15];int ans;//min grid number int sum;//all integers and half int cnt;//current lattice number int xx[4] = {-1, 0, 1, 0}, Yy[4 ] = {0, 1, 0, -1};void dfs (int x, int y, int cnt, int value) {if (x < 1 | | y < 1 | | x > N | | y > M) return;if (v Alue = = sum) {if (cnt < ans) ans = cnt;return;} else if (value > Sum) return;else {for (int i = 0; i < 4; i++) {int p = x + xx[i], q = y + yy[i];if (!vis[p][q]) {vis[p ][Q] = 1;dfs (p, Q, Cnt+1, value+a[p][q]); Vis[p][q] = 0;}}} int main () {while (scanf ("%d%d", &m, &n)! = EOF) {sum = 0;for (int i = 1; I <= n; i++) {for (int j = 1; J <= M ; J + +) {scanf ("%d", &a[i][j]); Vis[i][j] = 0;sum + = A[i][j];}} Sum/= 2;ans = inf;vis[1][1] = 1;dfs (1, 1, 1, a[1][1]), if (ans = = INF) printf ("0\n"), Else printf ("%d\n", ans); }return 0;}

Blue Bridge Cup-cut lattice (simple Dfs)