119 lines
3.5 KiB
C++
119 lines
3.5 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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// dijkstra算法模板及其用法
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// https://www.cnblogs.com/yoyo-sincerely/p/6400906.html
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// https://www.cnblogs.com/mycapple/archive/2012/08/12/2634227.html
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/***************************************
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* About: 有向图的Dijkstra算法实现
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* Author: Tanky Woo
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* Blog: www.WuTianQi.com
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***************************************/
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const int maxnum = 100;
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const int maxint = 999999;
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// 各数组都从下标1开始
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int dist[maxnum]; // 表示当前点到源点的最短路径长度
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int prevArr[maxnum]; // 记录当前点的前一个结点
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int c[maxnum][maxnum]; // 记录图的两点间路径长度
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int n, line; // 图的结点数和路径数
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void Dijkstra(int n, int v, int *dist, int *prev, int c[maxnum][maxnum]) {
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bool s[maxnum]; // 判断是否已存入该点到S集合中
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for (int i = 1; i <= n; ++i) {
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dist[i] = c[v][i];
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s[i] = 0; // 初始都未用过该点
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if (dist[i] == maxint)
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prev[i] = 0;
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else
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prev[i] = v;
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}
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dist[v] = 0;
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s[v] = 1;
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// 依次将未放入S集合的结点中，取dist[]最小值的结点，放入结合S中
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// 一旦S包含了所有V中顶点，dist就记录了从源点到所有其他顶点之间的最短路径长度
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for (int i = 2; i <= n; ++i) {
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int tmp = maxint;
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int u = v;
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// 找出当前未使用的点j的dist[j]最小值
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for (int j = 1; j <= n; ++j)
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if ((!s[j]) && dist[j] < tmp) {
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u = j; // u保存当前邻接点中距离最小的点的号码
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tmp = dist[j];
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}
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s[u] = 1; // 表示u点已存入S集合中
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// 更新dist
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for (int j = 1; j <= n; ++j)
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if ((!s[j]) && c[u][j] < maxint) {
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int newdist = dist[u] + c[u][j];
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if (newdist < dist[j]) {
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dist[j] = newdist;
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prev[j] = u;
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}
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}
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}
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}
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void searchPath(int *prev, int v, int u) {
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int que[maxnum];
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int tot = 1;
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que[tot] = u;
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tot++;
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int tmp = prev[u];
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while (tmp != v) {
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que[tot] = tmp;
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tot++;
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tmp = prev[tmp];
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}
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que[tot] = v;
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for (int i = tot; i >= 1; --i)
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if (i != 1)
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cout << que[i] << " -> ";
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else
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cout << que[i] << endl;
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}
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int main() {
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freopen("../MiniPath/Dijkstra2.txt", "r", stdin);
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// 各数组都从下标1开始
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// 输入结点数
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cin >> n;
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// 输入路径数
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cin >> line;
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int p, q, len; // 输入p, q两点及其路径长度
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// 初始化c[][]为maxint
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for (int i = 1; i <= n; ++i)
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for (int j = 1; j <= n; ++j)
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c[i][j] = maxint;
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for (int i = 1; i <= line; ++i) {
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cin >> p >> q >> len;
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if (len < c[p][q]) // 有重边
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{
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c[p][q] = len; // p指向q
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c[q][p] = len; // q指向p，这样表示无向图
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}
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}
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for (int i = 1; i <= n; ++i)
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dist[i] = maxint;
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for (int i = 1; i <= n; ++i) {
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for (int j = 1; j <= n; ++j)
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printf("%8d", c[i][j]);
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printf("\n");
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}
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Dijkstra(n, 1, dist, prevArr, c);
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// 最短路径长度
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cout << "源点到最后一个顶点的最短路径长度: " << dist[n] << endl;
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// 路径
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cout << "源点到最后一个顶点的路径为: ";
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searchPath(prevArr, 1, n);
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} |