#include "stdio.h" void main() { long int money[6][6] = { {0,56,35,21,51,60}, {56,0,21,57,78,70},{35,21,0,36,68,0}, {21,57,36,0,51,61},{51,78,68,51,0,13}, {60,70,0,61,13,0} long int sta2,sta3,sta4,sta5,sta6,i =0,x =0,y = 0,min = O,xuhao = 0; long int all[10000]12,13,14,15,16,17,18; printf("all case:\n"); for (sta2 = 1; sta2 <= 7; sta2++)for (sta3 = 1; sta3 <= 7; sta3++) for (sta4 = 1; sta4 <= 7; sta4++)for (sta5 = 1; sta5 <= 7; sta5++) for (sta6 = 1; sta6 <= 7; sta6++) if (sta3 != sta2 && sta4 != sta3 && sta4 != sta2 && sta5 != sta4 && sta5 != sta3 && sta5 != sta2 && sta6 != sta5 && sta6 != sta4 && sta6 != sta3 && sta6 != sta2 ) { all[i] = money[0][sta2] + money[sta2][sta3] + money[sta3][sta4] + money[sta4][sta5] + money[sta5][sta6] + money[sta6][0] ; i++; { if (i == 4039) { l2 = sta2; l3 = sta3; l4 = sta4; l5 = sta5; l6 = sta6; }} for (x = 0; x < ; x++) { printf("%5d,"all[x]); if (x != 0 && x % 10 == 0) printf("\n"); printf("\n"); min = all[0]; for (y = 1; y < i; y++) { if (all[y] <= min) { min = all[y]; xuhao = y; } } printf("min=%3d, xuhao=%3d\n", min, xuhao); printf("best way is:0,%3d, %3d, %3d, %3d, %3d\n", l2, l3, l4, l5, l6); }

这段代码实现了一个旅行商问题的解法，通过枚举所有可能的路径，计算出每个路径的总花费，并找到花费最小的路径。其中，money数组表示每两个城市之间的距离，sta2~sta6表示旅行的城市序号，i表示路径数量，all数组存储所有路径的总花费，min表示最小花费，xuhao表示对应路径的序号，l2~l6表示最优路径经过的城市序号。不过，这段代码存在一些问题，比如变量未定义、数组下标越界等，需要进行修正才能运行。

相关问题

帮我修改代码#include "stdio.h" void main() { long int money[6][6] = { {0,56,35,21,51,60}, {56,0,21,57,78,70},{35,21,0,36,68,0}, {21,57,36,0,51,61},{51,78,68,51,0,13}, {60,70,0,61,13,0} long int sta2,sta3,sta4,sta5,sta6,i =0,x =0,y = 0,min = O,xuhao = 0; long int all[10000]12,13,14,15,16,17,18; printf("all case:\n"); for (sta2 = 1; sta2 <= 7; sta2++)for (sta3 = 1; sta3 <= 7; sta3++) for (sta4 = 1; sta4 <= 7; sta4++)for (sta5 = 1; sta5 <= 7; sta5++) for (sta6 = 1; sta6 <= 7; sta6++) if (sta3 != sta2 && sta4 != sta3 && sta4 != sta2 && sta5 != sta4 && sta5 != sta3 && sta5 != sta2 && sta6 != sta5 && sta6 != sta4 && sta6 != sta3 && sta6 != sta2 ) { all[i] = money[0][sta2] + money[sta2][sta3] + money[sta3][sta4] + money[sta4][sta5] + money[sta5][sta6] + money[sta6][0] ; i++; { if (i == 4039) { l2 = sta2; l3 = sta3; l4 = sta4; l5 = sta5; l6 = sta6; }} for (x = 0; x < ; x++) { printf("%5d,"all[x]); if (x != 0 && x % 10 == 0) printf("\n"); printf("\n"); min = all[0]; for (y = 1; y < i; y++) { if (all[y] <= min) { min = all[y]; xuhao = y; } } printf("min=%3d, xuhao=%3d\n", min, xuhao); printf("best way is:0,%3d, %3d, %3d, %3d, %3d\n", l2, l3, l4, l5, l6); }

修改后的代码如下：

#include <stdio.h>

int main() {
long int money[6][6] = { {0,56,35,21,51,60}, {56,0,21,57,78,70}, {35,21,0,36,68,0}, {21,57,36,0,51,61}, {51,78,68,51,0,13}, {60,70,0,61,13,0} };
long int sta2, sta3, sta4, sta5, sta6, i = 0, x = 0, y = 0, min = money[0][1]+money[1][2]+money[2][3]+money[3][4]+money[4][5]+money[5][0], xuhao = 0;
long int all[10000];

printf("all cases:\n");

for (sta2 = 1; sta2 <= 5; sta2++)
for (sta3 = 1; sta3 <= 5; sta3++)
for (sta4 = 1; sta4 <= 5; sta4++)
for (sta5 = 1; sta5 <= 5; sta5++)
for (sta6 = 1; sta6 <= 5; sta6++)
if (sta2 != sta3 && sta2 != sta4 && sta2 != sta5 && sta2 != sta6 &&
sta3 != sta4 && sta3 != sta5 && sta3 != sta6 &&
sta4 != sta5 && sta4 != sta6 &&
sta5 != sta6) {
all[i] = money[0][sta2] + money[sta2][sta3] + money[sta3][sta4] +
money[sta4][sta5] + money[sta5][sta6] + money[sta6][0];
i++;
}

for (x = 0; x < i; x++) {
printf("%5d,", all[x]);
if (x != 0 && x % 10 == 0)
printf("\n");
}

printf("\n");

for (y = 0; y < i; y++) {
if (all[y] <= min) {
min = all[y];
xuhao = y;
}
}

sta2 = (xuhao / 24) + 1;
sta3 = (xuhao % 24 / 6) + 1;
sta4 = (xuhao % 6 / 2) + 1;
sta5 = (xuhao % 2 == 0) ? 2 : 3;
sta6 = (xuhao % 2 == 0) ? 3 : 2;

printf("min=%3d, xuhao=%3d\n", min, xuhao);
printf("best way is:0,%3d, %3d, %3d, %3d, %3d\n", sta2, sta3, sta4, sta5, sta6);

return 0;
}

这个程序的功能是求解旅行商问题的最优解，即给定N个城市之间的距离，求出经过所有城市的最短路径。程序中的money数组表示城市之间的距离，其中money[i][j]表示从城市i到城市j的距离。程序利用6重循环枚举所有可能的路径，计算每一条路径的总距离，并保存到all数组中。然后遍历all数组，找出最小值及其对应的路径。最后输出最小值及其对应的路径。

#include"stdio.h" #define _CRT_SECURE_NO_WARNINGS #include <cstdio> void main() { long int money[8][8] = { {0,56,35,21,51,60,43,39}, {56,0,21,57,78,70,64,49},{35,21,0,36,68,0,70,60}, {21,57,36,0,51,61,65,26},{51,78,68,51,0,13,45,62}, {60,70,0,61,13,0,53,26}, {43,64, 70,65,45,53,0,50}, {39,49,60,26,62,26, 50,0} }; long int sta2, sta3, sta4, sta5, sta6, sta7, sta8, i = 0, x = 0, y = 0, min = 0, xuhao = 0; long int all[10000], l2=0, l3=0, l4=0, l5=0, l6=0, l7=0, l8=0; printf("all case:\n"); for (sta2 = 1; sta2 <= 7; sta2++) for (sta3 = 1; sta3 <= 7; sta3++) for (sta4 = 1; sta4 <= 7; sta4++) for (sta5 = 1; sta5 <= 7; sta5++) for (sta6 = 1; sta6 <= 7; sta6++) for (sta7 = 1; sta7 <= 7; sta7++) for (sta8 = 1; sta8 <= 7; sta8++) if (sta3 != sta2 && sta4 != sta3 && sta4 != sta2 && sta5 != sta4 && sta5 != sta3 && sta5 != sta2 && sta6 != sta5 && sta6 != sta4 && sta6 != sta3 && sta6 != sta2 && sta7 != sta6 && sta7 != sta5 && sta7 != sta4 && sta7 != sta3 && sta7 != sta2 && sta8 != sta7 && sta8 != sta6 && sta8 != sta5 && sta8 != sta4 && sta8 != sta3 && sta8 != sta2) { all[i] = money[0][sta2] + money[sta2][sta3] + money[sta3][sta4] + money[sta4][sta5] + money[sta5][sta6] + money[sta6][sta7] + money[sta7][sta8] + money[sta8][0]; i++; } if (i == 4039) { l2 = sta2; l3 = sta3; l4 = sta4; l5 = sta5; l6 = sta6; l7 = sta7; l8 = sta8; } } for (x = 0; x < i; x++){ { printf("%5d%", all[x]); if (x != 0 && x % 10 == 0) printf("\n"); printf("\n"); min = all[0]; for (y = 1; y < i; y++) { if (all[y] <= min) { min = all[y]; xuhao = y; } } printf("min=%3d, xuhao=%3d\n", min, xuhao); printf("best way is:0,%3d, %3d, %3d, %3d, %3d,% 3d,%3d\n", l2, l3, l4, l5, l6, l7, l8); }

根据您提供的代码，我发现其中存在一些错误：

1. `void main()` 应该改为 `int main()`，因为 `main()` 函数应该返回一个整型值。
2. 在 `printf("%5d%", all[x]);` 这行代码中，`%` 符号应该只有一个，表示格式控制符的开始和结束，应该改为 `printf("%5d", all[x]);`。
3. 在 `printf("best way is:0,%3d, %3d, %3d, %3d, %3d,% 3d,%3d\n", l2, l3, l4, l5, l6, l7, l8);` 这行代码中，`% 3d` 中的空格不应该存在，应该改为 `%3d`。

另外，您的代码中存在一些逻辑错误。在第一个 `for` 循环结束后，`sta2` 到 `sta8` 的值都将是 8，因此在最后的 `if` 语句中判断 `i` 是否等于 4039 的操作将永远不会被执行。您需要将最后的 `if` 语句放置在第一个 `for` 循环的内部，以便在每次迭代结束后判断条件。

修改后的代码如下所示：

#include <cstdio>

#define _CRT_SECURE_NO_WARNINGS

int main() {
    long int money[8][8] = {
        {0, 56, 35, 21, 51, 60, 43, 39},
        {56, 0, 21, 57, 78, 70, 64, 49},
        {35, 21, 0, 36, 68, 0, 70, 60},
        {21, 57, 36, 0, 51, 61, 65, 26},
        {51, 78, 68, 51, 0, 13, 45, 62},
        {60, 70, 0, 61, 13, 0, 53, 26},
        {43, 64, 70, 65, 45, 53, 0, 50},
        {39, 49, 60, 26, 62, 26, 50, 0}
    };

    long int sta2, sta3, sta4, sta5, sta6, sta7, sta8, i = 0, x = 0, y = 0, min = 0, xuhao = 0;
    long int all[10000], l2 = 0, l3 = 0, l4 = 0, l5 = 0, l6 = 0, l7 = 0, l8 = 0;

    printf("all case:\n");
    for (sta2 = 1; sta2 <= 7; sta2++) {
        for (sta3 = 1; sta3 <= 7; sta3++) {
            for (sta4 = 1; sta4 <= 7; sta4++) {
                for (sta5 = 1; sta5 <= 7; sta5++) {
                    for (sta6 = 1; sta6 <= 7; sta6++) {
                        for (sta7 = 1; sta7 <= 7; sta7++) {
                            for (sta8 = 1; sta8 <= 7; sta8++) {
                                if (sta3 != sta2 &&
                                    sta4 != sta3 && sta4 != sta2 &&
                                    sta5 != sta4 && sta5 != sta3 && sta5 != sta2 &&
                                    sta6 != sta5 && sta6 != sta4 && sta6 != sta3 && sta6 != sta2 &&
                                    sta7 != sta6 && sta7 != sta5 && sta7 != sta4 && sta7 != sta3 && sta7 != sta2 &&
                                    sta8 != sta7 && sta8 != sta6 && sta8 != sta5 && sta8 != sta4 && sta8 != sta3 && sta8 != sta2) {
                                    all[i] = money[0][sta2] + money[sta2][sta3] + money[sta3][sta4] +
                                        money[sta4][sta5] + money[sta5][sta6] + money[sta6][sta7] + money[sta7][sta8] + money[sta8][0];
                                    i++;
                                    if (i == 4039) {
                                        l2 = sta2;
                                        l3 = sta3;
                                        l4 = sta4;
                                        l5 = sta5;
                                        l6 = sta6;
                                        l7 = sta7;
                                        l8 = sta8;
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }

    for (x = 0; x < i; x++) {
        printf("%5d", all[x]);
        if (x != 0 && x % 10 == 0) {
            printf("\n");
        }
    }

    printf("\nmin=%3d, xuhao=%3d\n", all[0], 0);
    for (y = 1; y < i; y++) {
        if (all[y] <= min) {
            min = all[y];
            xuhao = y;
        }
    }
    printf("min=%3d, xuhao=%3d\n", min, xuhao);
    printf("best way is:0,%3d, %3d, %3d, %3d, %3d,%3d,%3d\n", l2, l3, l4, l5, l6, l7, l8);

    return 0;
}

希望以上信息能够帮助到您！