如何获得在阵列/矩阵的1的线路最长

本文介绍了如何获得在阵列/矩阵的1的线路最长的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我的任务是找到阵列1的的最长的路线。 Hoizontal和垂直​​。阵列仅由0和1的，看起来如这样的：

4 40 1 1 10 1 0 10 1 1 01 0 1 0

输出应打印[i] [j]的首发1，和[I] [J]。收官1，所以水平应该是[1] [0] [3] [0]。

我用我的GETCOLOR（）函数来得到[I] [J]现货价值。

我想这个对于WAAAAY长，差不多花了整整一个星期的this.I有一些想法，但没有奏效。也许是因为我是新来的C和完全新的数组。

我知道我应该去通过阵列，每1发现它应该保存坐标开始，然后转到下一个，每节约1发现结束。 0被发现后，比较lenght，如果lenght是国内规模最大，覆盖lenght。但我没写code正确。有人能帮助我的广告撰写code？非常感谢你。

编辑：这是我的，但我在一开始我只是，它不工作已

如果（ARR！= NULL）{  而（J＆LT; arr-＆GT; COLS）{    而（I＆LT; = arr-＆GT;行）{            如果（GETCOLOR（ARR，I，J）== 1）{                startI = I;                startJ = j的;                打破;            }            其他            我++;    }    I = 0;    而（I＆LT; = arr-＆GT;行）{            如果（GETCOLOR（ARR，I，J）== 1）{                恩迪= I;                endJ = j的;            }            我++;    }    I = 0;   的printf（开始％D \\ NEND％D \\ n第％d个\\ n \\ n，startI，startJ，恩迪，endJ，线）;   J ++;   }}

解决方案

对不起，悔之晚矣。我没有写你到底怎么想的那样，但是这会真的帮助你。只要阅读它，你会得到的想法。

链接到code pastebin/vLATASab

或者查看这里：

的#include＆LT;＆stdio.h中GT;主要（）{    INT ARR [4] [4];    改编[0] [0] = 0;常用3 [0] [1] = 1;常用3 [0] [2] = 1;常用3 [0] [3] = 1;    改编[1] [0] = 0;常用3 [1] [1] = 1;常用3 [1] [2] = 0;常用3 [1] [3] = 1;    ARR [2] [0] = 0;常用3 [2] [1] = 1;常用3 [2] [2] = 1;常用3 [2] [3] = 0;    改编[3] [0] = 1;常用3 [3] [1] = 0;常用3 [3] [2] = 1;常用3 [3] [3] = 1;    INT I，J，K;    INT线，LINE_START，LINE_END，LINE_MAX = 0;    INT关口，col_start，co​​l_end，col_max = 0;    // Horizently    为（K = 0; K＆4;; k ++）    {        对于（I = 0; I＆下; 4;我+ +）        {            如果（ARR [k]的[I]！）继续;            为（J =; J＆下; 4; J ++）            {                如果突破（ARR [K] [J]！）;            }            j--;            如果（J-I + 1＆GT; LINE_MAX）            {                行= K;                LINE_START = I;                LINE_END = j的;                LINE_MAX = LINE_END-LINE_START + 1;            }        }    }    的printf（horizo​​ntaly \\ n）;    的printf（启动：[％D]。[％D]。\\ n，行，LINE_START）;    的printf（结束：[％D]。[％D]。\\ n，行，LINE_END）;    // Verticaly    为（K = 0; K＆4;; k ++）    {        对于（I = 0; I＆下; 4;我+ +）        {            如果（ARR [I] [K]！）继续;            为（J =; J＆下; 4; J ++）            {                如果突破（ARR [J] [K]！）;            }            j--;            如果（J-I + 1＆GT; col_max）            {                COL = K;                col_start = I;                col_end = j的;                col_max = col_end-col_start + 1;            }        }    }    的printf（\\ nverticaly \\ n）;    的printf（启动：[％D]。[％D]。\\ n，col_start，列）;    的printf（结束：[％D]。[％D]。\\ n，col_end，列）;}

My task is to find "longest line" of 1's in array. Hoizontal and Vertical. Array is made only of 0's and 1's, and looks for example like this:

4 4 0 1 1 1 0 1 0 1 0 1 1 0 1 0 1 0

Output should print [ i ][ j ] of "starting" 1, and [ i ][ j ] of "ending" 1., So horizontal should be [1][0] [3][0].

I am using my getcolor() function to get value on [ i ][ j ] spot.

I am thinking about this for WAAAAY to long, spent almost whole week on this.I had some ideas, but none worked. Maybe because I am new to C and completely new to arrays.

I know that I should go through the array, on every 1 found it should save the coordinates to "start", and go to next, saving every 1 found to "end". After 0 is found, compare the lenght, and if the lenght is largest, overwrite the lenght. But I didn't manage to write the code right. Can someone help me ad write the code? Thank you very much.

Edit: This is what I have, but I am just at the start and it doesn't work already:

if(arr != NULL) { while (j < arr->cols) { while (i <=arr->rows) { if (getcolor(arr, i, j) == 1) { startI = i; startJ = j; break; } else i++; } i=0; while (i <=arr->rows) { if (getcolor(arr, i, j) == 1) { endI = i; endJ = j; } i++; } i=0; printf("start %d %d\nend %d %d\nline %d\n\n", startI, startJ, endI, endJ, line); j++; } }

解决方案

Sorry for being too late. I didn't write it exactly how you want it, but this will realy help you. Just read it and you will get the idea.

Link to code pastebin/vLATASab

Or view it here:

#include <stdio.h> main() { int arr[4][4]; arr[0][0] = 0;arr[0][1] = 1;arr[0][2] = 1;arr[0][3] = 1; arr[1][0] = 0;arr[1][1] = 1;arr[1][2] = 0;arr[1][3] = 1; arr[2][0] = 0;arr[2][1] = 1;arr[2][2] = 1;arr[2][3] = 0; arr[3][0] = 1;arr[3][1] = 0;arr[3][2] = 1;arr[3][3] = 1; int i, j, k; int line, line_start, line_end, line_max = 0; int col, col_start, col_end, col_max = 0; // Horizently for (k=0; k<4; k++) { for (i=0; i<4; i++) { if (!arr[k][i]) continue; for (j=i; j<4; j++) { if (!arr[k][j]) break; } j--; if (j-i+1>line_max) { line = k; line_start = i; line_end = j; line_max = line_end-line_start+1; } } } printf("horizontaly\n"); printf("start: [%d][%d]\n", line, line_start); printf("end: [%d][%d]\n", line, line_end); // Verticaly for (k=0; k<4; k++) { for (i=0; i<4; i++) { if (!arr[i][k]) continue; for (j=i; j<4; j++) { if (!arr[j][k]) break; } j--; if (j-i+1>col_max) { col = k; col_start = i; col_end = j; col_max = col_end-col_start+1; } } } printf("\nverticaly\n"); printf("start: [%d][%d]\n", col_start, col); printf("end: [%d][%d]\n", col_end, col); }