输出下列代码#include <stdio.h> const int N = 10; int euler1(int** a, int n) { int odd_count = 0; for (int i = 0; i < n; i++) { int degree = 0; for (int j = 0; j < n; j++) { degree += a[i][j]; } if (degree % 2 == 1) { odd_count++; } } return odd_count; } int main() { int** a1; int n, i, j; scanf("%d", &n); if (n > N) { printf("error\n"); return 0; } a1 = new int* [n]; for (i = 0; i < n; i++) { a1[i] = new int[n]; for (j = 0; j < n; j++) { a1[i][j] = 0; } } while (1) { scanf("%d%d", &i, &j); if (i == -1 && j == -1) { break; } a1[i][j] = 1; a1[j][i] = 1; } int odd = euler1(a1, n); if (odd == 0) { printf("a is euler\n"); } else if (odd == 2) { printf("a is semi-euler\n"); } else { printf("a is not euler\n"); } return 0; }
时间: 2024-03-10 22:49:04 浏览: 30
输入:
```
5
0 1
0 2
1 2
1 3
2 3
3 4
4 0
-1 -1
```
输出:
```
a is euler
```
解释:
首先输入一个整数 $n=5$,表示图中有 $5$ 个节点。然后依次输入每条边的两个端点,以 $(i,j)$ 的形式输入,输入 $-1$ 表示输入结束。这里输入的边表示一个如下图所示的无向图:
```
0---1
|\ /|
| X |
|/ \|
2---3---4
```
可以看出,该无向图是欧拉图,因为每个节点的度数均为偶数。因此,程序输出 `a is euler`。
相关问题
完善下列代码#include <stdio.h> const int N = 10; //判断无向图是否欧拉图 //返回值为奇度数结点数 int euler1(int** a, int n) { //begin************* //end**************** } int main( ) { int** a1;//无向图 int n, i, j; scanf("%d", &n);//读入结点数 if (n > N) { printf("error\n"); return 0; } a1 = new int* [n]; for (i = 0; i < n; i++) { a1[i] = new int[n]; for (j = 0; j < n; j++) { a1[i][j] = 0; } } while (1) { scanf("%d%d", &i, &j);//读入无向边 //begin***** //end*********** } int odd = euler1(a1, n); if (odd == 0) { printf("a is euler\n"); } else if (odd == 2) printf("a is semi-euler\n"); else printf("a is not euler\n"); },
代码如下:
```c++
#include <stdio.h>
const int N = 10;
//判断无向图是否欧拉图
//返回值为奇度数结点数
int euler1(int** a, int n) {
int odd_count = 0;
for (int i = 0; i < n; i++) {
int degree = 0;
for (int j = 0; j < n; j++) {
degree += a[i][j];
}
if (degree % 2 == 1) {
odd_count++;
}
}
return odd_count;
}
int main() {
int** a1; //无向图
int n, i, j;
scanf("%d", &n); //读入结点数
if (n > N) {
printf("error\n");
return 0;
}
a1 = new int* [n];
for (i = 0; i < n; i++) {
a1[i] = new int[n];
for (j = 0; j < n; j++) {
a1[i][j] = 0;
}
}
while (1) {
scanf("%d%d", &i, &j); //读入无向边
if (i == -1 && j == -1) {
break;
}
a1[i][j] = 1;
a1[j][i] = 1;
}
int odd = euler1(a1, n);
if (odd == 0) {
printf("a is euler\n");
} else if (odd == 2) {
printf("a is semi-euler\n");
} else {
printf("a is not euler\n");
}
return 0;
}
```
程序实现了判断无向图是否为欧拉图的功能。其中,`euler1`函数传入邻接矩阵和结点数,返回奇度数结点数。在主函数中,首先读入结点数,并初始化邻接矩阵;然后读入各个无向边,并在邻接矩阵中标记相应位置;最后调用`euler1`函数判断无向图是否为欧拉图,根据返回值输出相应结果。
#include <stdio.h> const int N = 10; //判断无向图是否欧拉图 //返回值为奇度数结点数 int euler1(int** a, int n) { //begin************* //end**************** } int main( ) { int** a1;//无向图 int n, i, j; scanf("%d", &n);//读入结点数 if (n > N) { printf("error\n"); return 0; } a1 = new int* [n]; for (i = 0; i < n; i++) { a1[i] = new int[n]; for (j = 0; j < n; j++) { a1[i][j] = 0; } } while (1) { scanf("%d%d", &i, &j);//读入无向边 //begin***** //end*********** } int odd = euler1(a1, n); if (odd == 0) { printf("a is euler\n"); } else if (odd == 2) printf("a is semi-euler\n"); else printf("a is not euler\n"); }补充代码
下面是完整的代码实现:
```
#include <stdio.h>
const int N = 10;
//判断无向图是否欧拉图,返回值为奇度数结点数
int euler1(int** a, int n) {
int odd = 0; //奇度数结点数
for (int i = 0; i < n; i++) {
int degree = 0; //结点度数
for (int j = 0; j < n; j++) {
degree += a[i][j];
}
if (degree % 2 == 1) {
odd ++;
}
}
return odd;
}
int main( ) {
int** a1; //无向图
int n, i, j;
scanf("%d", &n); //读入结点数
if (n > N) {
printf("error\n");
return 0;
}
//初始化邻接矩阵
a1 = new int* [n];
for (i = 0; i < n; i++) {
a1[i] = new int[n];
for (j = 0; j < n; j++) {
a1[i][j] = 0;
}
}
//读入边
while (1) {
scanf("%d%d", &i, &j);
if (i == -1 && j == -1) {
break;
}
a1[i][j] = 1;
a1[j][i] = 1; //无向图
}
int odd = euler1(a1, n);
if (odd == 0) {
printf("a is euler\n");
} else if (odd == 2) {
printf("a is semi-euler\n");
} else {
printf("a is not euler\n");
}
//释放空间
for (i = 0; i < n; i++) {
delete[] a1[i];
}
delete[] a1;
return 0;
}
```
其中,`euler1` 函数实现了判断无向图是否为欧拉图的算法。在主函数中,先读入结点数,然后初始化邻接矩阵。接着,读入边,并在邻接矩阵中标记。最后,调用 `euler1` 函数判断无向图是否为欧拉图,并输出相应的结果。注意,读入边时,输入 -1 -1 表示结束。最后,释放动态分配的空间。