dijkstra构建最小生成树
时间: 2024-04-21 17:21:02 浏览: 10
Dijkstra算法是一种用于求解单源最短路径问题的经典算法,而不是用于构建最小生成树的算法。最小生成树是指在一个连通无向图中,找到一个包含所有顶点且边权重之和最小的树。
如果你想了解如何构建最小生成树,可以使用Prim算法或者Kruskal算法。下面是对这两种算法的简要介绍:
1. Prim算法:
- Prim算法是一种贪心算法,用于构建最小生成树。
- 算法从一个起始顶点开始,逐步扩展生成树,直到包含所有顶点为止。
- 在每一步中,选择与当前生成树相连的边中权重最小的边,并将其加入生成树。
- 重复上述步骤,直到生成树包含所有顶点。
相关问题
构建图的最小生成树或求最短路径C语言
最小生成树算法:
使用Prim算法实现:
```c
#include <stdio.h>
#include <stdbool.h> // bool类型需要用到
#define MAX_VERTICES 100
#define INF 1000000
int num_vertices; // 图中顶点个数
int adj_matrix[MAX_VERTICES][MAX_VERTICES]; // 图的邻接矩阵
void prim(int start_vertex) {
bool visited[MAX_VERTICES] = {false}; // 顶点是否已被访问
int distance[MAX_VERTICES]; // 到各顶点的距离
int parent[MAX_VERTICES]; // 生成树的父节点
// 初始化
for (int i = 0; i < num_vertices; i++) {
visited[i] = false;
distance[i] = INF;
parent[i] = -1;
}
distance[start_vertex] = 0;
for (int i = 0; i < num_vertices - 1; i++) { // 循环n-1次,构建n-1条边
int min_distance = INF;
int u; // 最小距离对应的顶点编号
// 找到未被访问过的、到起点距离最短的顶点
for (int j = 0; j < num_vertices; j++) {
if (!visited[j] && distance[j] < min_distance) {
min_distance = distance[j];
u = j;
}
}
visited[u] = true; // 对该顶点进行标记
// 更新与该顶点相邻的顶点到起点的距离
for (int v = 0; v < num_vertices; v++) {
if (adj_matrix[u][v] != 0 && !visited[v] && adj_matrix[u][v] < distance[v]) {
distance[v] = adj_matrix[u][v];
parent[v] = u;
}
}
}
// 输出生成树
printf("生成树:\n");
for (int i = 1; i < num_vertices; i++) { // 从第二个顶点开始,依次输出每个顶点的父节点和边的权值
printf("%d - %d (%d)\n", parent[i], i, adj_matrix[i][parent[i]]);
}
}
int main() {
printf("请输入图中顶点个数:");
scanf("%d", &num_vertices);
printf("请输入图的邻接矩阵:\n");
for (int i = 0; i < num_vertices; i++) {
for (int j = 0; j < num_vertices; j++) {
scanf("%d", &adj_matrix[i][j]);
}
}
printf("请输入起点编号:");
int start_vertex;
scanf("%d", &start_vertex);
prim(start_vertex);
return 0;
}
```
使用Kruskal算法实现:
```c
#include <stdio.h>
#include <stdbool.h> // bool类型需要用到
#define MAX_VERTICES 100
#define INF 1000000
int num_vertices; // 图中顶点个数
int adj_matrix[MAX_VERTICES][MAX_VERTICES]; // 图的邻接矩阵
int parent[MAX_VERTICES]; // 并查集数组
// 并查集查找根节点
int find(int x) {
while (parent[x] != -1) {
x = parent[x];
}
return x;
}
void kruskal() {
int total_weight = 0; // 最小生成树的总权值
int edge_count = 0; // 已选取的边的数量
// 初始化并查集
for (int i = 0; i < num_vertices; i++) {
parent[i] = -1;
}
// 对所有边按权值从小到大排序
struct {
int u, v, weight;
} edges[num_vertices * num_vertices]; // 边的数组,最多有n^2条边
int count = 0; // 实际边的数量
for (int i = 0; i < num_vertices; i++) {
for (int j = i + 1; j < num_vertices; j++) { // 使用邻接矩阵,只需要遍历矩阵的上三角
if (adj_matrix[i][j] != 0) {
edges[count].u = i;
edges[count].v = j;
edges[count].weight = adj_matrix[i][j];
count++;
}
}
}
for (int i = 0; i < count - 1; i++) { // 冒泡排序
for (int j = 0; j < count - i - 1; j++) {
if (edges[j].weight > edges[j+1].weight) {
struct {
int u, v, weight;
} temp = edges[j];
edges[j] = edges[j+1];
edges[j+1] = temp;
}
}
}
// 依次选取权值最小的边,如果两个顶点不在同一个连通分量中,则加入最小生成树
printf("生成树:\n");
while (edge_count < num_vertices - 1) { // 已选取的边的数量小于n-1时,继续选取
int u = edges[edge_count].u;
int v = edges[edge_count].v;
int weight = edges[edge_count].weight;
int root_u = find(u);
int root_v = find(v);
if (root_u != root_v) { // 如果两个顶点不在同一个连通分量中
parent[root_u] = root_v; // 合并连通分量
printf("%d - %d (%d)\n", u, v, weight); // 输出边
total_weight += weight; // 加上边的权值
}
edge_count++;
}
printf("最小权值和:%d\n", total_weight);
}
int main() {
printf("请输入图中顶点个数:");
scanf("%d", &num_vertices);
printf("请输入图的邻接矩阵:\n");
for (int i = 0; i < num_vertices; i++) {
for (int j = 0; j < num_vertices; j++) {
scanf("%d", &adj_matrix[i][j]);
}
}
kruskal();
return 0;
}
```
最短路径算法:
使用Dijkstra算法实现:
```c
#include <stdio.h>
#include <stdbool.h> // bool类型需要用到
#define MAX_VERTICES 100
#define INF 1000000
int num_vertices; // 图中顶点个数
int adj_matrix[MAX_VERTICES][MAX_VERTICES]; // 图的邻接矩阵
void dijkstra(int start_vertex) {
bool visited[MAX_VERTICES] = {false}; // 顶点是否已被访问
int distance[MAX_VERTICES]; // 起点到各顶点的距离
// 初始化
for (int i = 0; i < num_vertices; i++) {
visited[i] = false;
distance[i] = INF;
}
distance[start_vertex] = 0;
for (int i = 0; i < num_vertices; i++) { // 循环n次,每次确定一个顶点的最短路径
int min_distance = INF;
int u; // 最小距离对应的顶点编号
// 找到未被访问过的、到起点距离最短的顶点
for (int j = 0; j < num_vertices; j++) {
if (!visited[j] && distance[j] < min_distance) {
min_distance = distance[j];
u = j;
}
}
visited[u] = true; // 对该顶点进行标记
// 更新与该顶点相邻的顶点到起点的距离
for (int v = 0; v < num_vertices; v++) {
if (adj_matrix[u][v] != 0 && !visited[v] && distance[u] + adj_matrix[u][v] < distance[v]) {
distance[v] = distance[u] + adj_matrix[u][v];
}
}
}
// 输出到各顶点的最短路径
printf("起点到各顶点的最短路径:\n");
for (int i = 0; i < num_vertices; i++) {
printf("%d -> %d: %d\n", start_vertex, i, distance[i]);
}
}
int main() {
printf("请输入图中顶点个数:");
scanf("%d", &num_vertices);
printf("请输入图的邻接矩阵:\n");
for (int i = 0; i < num_vertices; i++) {
for (int j = 0; j < num_vertices; j++) {
scanf("%d", &adj_matrix[i][j]);
}
}
printf("请输入起点编号:");
int start_vertex;
scanf("%d", &start_vertex);
dijkstra(start_vertex);
return 0;
}
```
使用Bellman-Ford算法实现:
```c
#include <stdio.h>
#include <stdbool.h> // bool类型需要用到
#define MAX_VERTICES 100
#define INF 1000000
int num_vertices; // 图中顶点个数
int adj_matrix[MAX_VERTICES][MAX_VERTICES]; // 图的邻接矩阵
void bellman_ford(int start_vertex) {
int distance[MAX_VERTICES]; // 起点到各顶点的距离
// 初始化
for (int i = 0; i < num_vertices; i++) {
distance[i] = INF;
}
distance[start_vertex] = 0;
for (int i = 0; i < num_vertices - 1; i++) { // 循环n-1次,每次松弛n-1条边
for (int u = 0; u < num_vertices; u++) {
for (int v = 0; v < num_vertices; v++) {
if (adj_matrix[u][v] != 0 && distance[u] + adj_matrix[u][v] < distance[v]) {
distance[v] = distance[u] + adj_matrix[u][v];
}
}
}
}
// 判断是否存在负权回路
bool has_negative_cycle = false;
for (int u = 0; u < num_vertices; u++) {
for (int v = 0; v < num_vertices; v++) {
if (adj_matrix[u][v] != 0 && distance[u] + adj_matrix[u][v] < distance[v]) {
has_negative_cycle = true;
break;
}
}
}
if (has_negative_cycle) {
printf("图中存在负权回路\n");
} else {
// 输出到各顶点的最短路径
printf("起点到各顶点的最短路径:\n");
for (int i = 0; i < num_vertices; i++) {
printf("%d -> %d: %d\n", start_vertex, i, distance[i]);
}
}
}
int main() {
printf("请输入图中顶点个数:");
scanf("%d", &num_vertices);
printf("请输入图的邻接矩阵:\n");
for (int i = 0; i < num_vertices; i++) {
for (int j = 0; j < num_vertices; j++) {
scanf("%d", &adj_matrix[i][j]);
}
}
printf("请输入起点编号:");
int start_vertex;
scanf("%d", &start_vertex);
bellman_ford(start_vertex);
return 0;
}
```
构建图的最小生成树或求最短路径。(二选一部分) (1)、用Prim法求连通图的最小生成树(邻接表) (2)、用Dijkstra算法求某一确定顶点到所有其它顶点的最短路径,请在网上寻找到这段话
好的,以下是用Prim算法求连通图的最小生成树的Python实现(使用邻接表表示图):
```python
import heapq
def prim(graph):
# 任选一个顶点作为起点
start = list(graph.keys())[0]
# 初始化距离字典、已知最短路径集合和边集合
distance = {vertex: float('inf') for vertex in graph}
distance[start] = 0
visited = set()
edges = []
# 堆优化的Prim算法
heap = [(0, start)]
while heap:
(dist, current_vertex) = heapq.heappop(heap)
# 如果当前顶点已经在已知最短路径集合中,跳过
if current_vertex in visited:
continue
visited.add(current_vertex)
# 将当前顶点与已知最短路径集合中的顶点之间的边加入边集合中
if current_vertex != start:
edges.append((distance[current_vertex], current_vertex, parent))
# 更新与当前顶点相邻的顶点的距离
for neighbor, weight in graph[current_vertex].items():
if neighbor not in visited:
if weight < distance[neighbor]:
distance[neighbor] = weight
heapq.heappush(heap, (weight, neighbor))
parent = current_vertex
# 返回边集合
return edges
```
其中,graph是邻接表表示的图。这个实现中使用了堆优化,可以有效地加速算法。
以下是用Dijkstra算法求某一确定顶点到所有其它顶点的最短路径的Python实现(使用邻接表表示图):
```python
import heapq
def dijkstra(graph, start):
# 初始化距离字典和已知最短路径集合
distance = {vertex: float('inf') for vertex in graph}
distance[start] = 0
visited = set()
# 堆优化的Dijkstra算法
heap = [(0, start)]
while heap:
(dist, current_vertex) = heapq.heappop(heap)
# 如果当前顶点已经在已知最短路径集合中,跳过
if current_vertex in visited:
continue
visited.add(current_vertex)
# 更新与当前顶点相邻的顶点的距离
for neighbor, weight in graph[current_vertex].items():
if neighbor not in visited:
new_distance = dist + weight
if new_distance < distance[neighbor]:
distance[neighbor] = new_distance
heapq.heappush(heap, (new_distance, neighbor))
# 返回距离字典
return distance
```
其中,graph是邻接表表示的图,start是起点。这个实现中使用了堆优化,可以有效地加速算法。