Kruskal和prime算法的类实现，图的遍历BFS算法。

一.图的遍历

#include<iostream>
#include<queue>
#include<vector>
using namespace std;
int n, m;  //行数和列数
const int maxn = 100;
char g[maxn][maxn];  //图
bool vis[maxn][maxn];  //访问标记数组,false表示点没有被访问过
int disx[4] = { 0,0,1,-1 };   //四个
int disy[4] = { 1,-1,0,0 };   //方向
int dx, dy;   //起点位置
int counter;  //黑瓷砖的数目
bool flag = false; //判断是否有起点
struct node {
	int x, y;
}p[maxn];
queue<node> que;
vector<node> vec;
void graph(int n, int m) {
	cout << "请输入行数和列数 ：";
	cin >> n >> m;
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			cin >> g[i][j];
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 1; j <= m; ++j) {
			if (g[i][j] == '@') {
				dx = i; dy = j;
				flag = true;
				break;
			}
		}
		if (flag == true) break;
	}
	if (flag) {
		cout << "起始点的坐标是";
		printf("( %d , %d )\n\n", dx, dy);
	}
	else cout << "没有起点" << endl;
}
void bfs() {
	vec.clear();
	que.empty();
	memset(vis, false, sizeof vis);
	graph(n, m);
	if (!flag) {
		counter = 0;
		cout << "黑瓷砖的数目是" << counter << endl;
	}
	else {
		counter = 1;
		node point;
		point.x = dx, point.y = dy;
		que.push(point);
		vis[point.x][point.y] = true;
		vec.push_back(point);
		while (!que.empty()) {
			node temp = que.front();
			node next;
			que.pop();
			for (int i = 0; i < 4; ++i) {
				int x = temp.x + disx[i];
				int y = temp.y + disy[i];
				next.x = x; next.y = y;
				if (g[next.x][next.y] == 'b'&&vis[next.x][next.y] == false) {
					vis[next.x][next.y] = true;
					que.push(next); ++counter;
					vec.push_back(next);
				}
			}
		}
		cout << "黑瓷砖的数目是" << counter << endl;
		cout << endl;
		int number = 0;
		cout << "且到达每个黑瓷砖的顺序是 ：" << endl;
		for (int i = 0; i < vec.size(); ++i) {
			if (i != vec.size() - 1) {
				++number;
				printf("( %d , %d)", vec[i].x, vec[i].y);
				cout << " -> ";
				if (number % 5 == 0) cout << endl;
			}
			else printf("( %d , %d)\n", vec[i].x, vec[i].y);
		}
	}
}
int main() {
	bfs();
	return 0;
}

糟糕的上机，prim似乎有bug

二.求无向连通图的生成树

#pragma once
#include<algorithm>
#include <fstream>
#include<iostream>
using namespace std;
const int inf = 1000000000;
int p[100];
int maze[100][100];
template<class T>
struct Edge
{
	int u, v;
	int net;
	T key;
	Edge() :u(-1), v(-1), key(0) {}
};
template<class T>
class MST {
protected:
	Edge<T>*edge;
	int maxSize, e;
	int head[100];
public:MST(int sz = 100) :maxSize(sz), e(0) {
	edge = new Edge<T>[sz];
	memset(head, -1, sizeof(head));
}
	   void addedge(Edge<T>& item);
	   void establish();
	   int Kruskal();
	   int Prim(int n);
};
template<class T>
void MST<T>::addedge(Edge<T>& item)
{
	int frm = item.u, to = item.v, w = item.key;
	edge[e].u = frm;
	edge[e].v = to;
	edge[e].key = w;
	edge[e].net = head[frm];
	head[frm] = e++;
	edge[e].u = to;
	edge[e].v = frm;
	edge[e].key = w;
	edge[e].net = head[to];
	head[to] = e++;
}
template<class T>
void MST<T>::establish()
{
	for (int i = 0; i < 100; i++)
		for (int j = 0; j< 100; j++)
			if (i == j)maze[i][j] = 0;
			else maze[i][j] = inf;
	int u, v, w;
	Edge<int> temp;
	while (cin >> u >> v >> w) {
		if (u == 0 && v == 0 && w == 0)break;
		temp.u = u, temp.v = v, temp.key = w;
		maze[u][v] = maze[v][u] = w;
		addedge(temp);
	}
	return;
}
bool cmp(Edge<int> a, Edge<int> b)
{
	return a.key < b.key;
}
int find(int x)
{
	return p[x] == x ? x : p[x] = find(p[x]);
}
template<class T>
int MST<T>::Kruskal()
{

	ofstream outfile;
	
	outfile.open("outdata.txt");
	outfile << "Kruskal计算如下:\n";
	int ans = 0;
	for (int i = 0; i < 50; i++)p[i] = i;
	sort(edge, edge + e, cmp);
	for (int i = 0; i < e; i++) {
		int x = find(edge[i].u);
		int y = find(edge[i].v);
		if (x != y) { ans += edge[i].key;
	//	cout << edge[i].u << " " << edge[i].v << endl;
		outfile << edge[i].u << "    " << edge[i].v<<"     权值：  " << edge[i].key << endl;

			  p[x] = y; }
	}
	outfile << "最小生成树中长度：   " << ans << endl;
	outfile.close();//关闭文件，保存文件。
	return ans;
}
Edge<int> dis[100];
int path[100];
template<class T>
int MST<T>::Prim(int n)
{
	
	bool vis[100];
	ofstream outfile;
	outfile.open("outdata.txt", ios::binary | ios::app | ios::in | ios::out);
	outfile << "Prim计算如下:\n\n\n";
	//int dis[100];
	for (int i = 0; i < 100; i++)
		dis[i].key = inf;
		memset(vis, 0, sizeof(vis));
		int ans = 0;
		dis[3].key = 0;
		while (1)
		{
			int k = 0;
			for (int j = 1; j <= n; j++)
			{
				if (!vis[j] && dis[j].key<dis[k].key)      //这一步找未收录顶点中dist值最小的
					k = j;            
			}
			if (!k) break;
			vis[k] = 1;                        
			ans += dis[k].key;
			if (dis[k].u > 0)
				outfile <<"    "<<dis[k].u << "    " << dis[k].v << "     权值" << dis[k].key << "\n" << endl;
			for (int j = 1; j <= n; j++)
			{
				if (dis[j].key>maze[k][j])
				{
					dis[j].key = maze[k][j];
					dis[j].u = k;
					dis[j].v = j;
					path[j] = k;
				}
			}
		}
		outfile << "总长度    " << ans << endl;
		outfile.close();//关闭文件，保存文件。
		
		return ans;
}

#include<iostream>
#include"graph.h"
using namespace std;
int main()
{
	MST<int> m;
	int n;
	cin >> n;
	m.establish();
	m.Kruskal();
	m.Prim(n);
	getchar();
	return 0;
}