Floyd算法計算最短路徑和迪傑斯特拉算法

#include<stdio.h>
int ans[101][101];
int main()
{
	int n, m;
	while (scanf("%d%d", &n, &m) != EOF)
	{
		if (m == 0 && n == 0)
			break;
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
				ans[i][j] = -1;
			ans[i][i] = 0;
		}
		for (int i = 0; i < m; i++)
		{
			int a, b, c;
			scanf("%d%d%d", &a, &b, &c);
			ans[a][b] = ans[b][a] = c;
		}
		for (int k = 1; k <= n; k++)
		{
			for (int i = 1; i <= n; i++)
			{
				for (int j = 1; j <= n; j++)
				{
					if (ans[i][k] == -1 || ans[k][j] == -1)
						continue;
					if (ans[i][j] == -1 || ans[i][k] + ans[k][j] < ans[i][j])
						ans[i][j] = ans[i][k] + ans[k][j];
				}
			}
		}
		printf("%d\n", ans[1][n]);
	}
	return 0;
}

#include<stdio.h>
#include<vector>
using namespace std;
struct list
{
	int next;
	int cost;
};
vector<list> edge[101];
bool mark[101];
int dis[101];
int main()
{
	int n, m;
	while (scanf("%d%d", &n, &m) != EOF)
	{
		if (m == 0 && n == 0)
			break;
		for (int i = 0; i <= n; i++)
			edge[i].clear();
		for (int i = 0; i < m; i++)
		{
			int a, b, c;
			scanf("%d%d%d", &a, &b, &c);
			list temp;
			temp.cost = c;
			temp.next = a;
			edge[b].push_back(temp);
			temp.next = b;
			edge[a].push_back(temp);
		}
		for (int i = 0; i <= n; i++)
		{
			mark[i] = false;
			dis[i] = -1;
		}
		dis[1] = 0;
		mark[1] = true;
		int newp = 1;
		for (int i = 0; i < n - 1; i++)
		{
			for (int j = 0; j < edge[newp].size(); j++)
			{
				int t = edge[newp][j].next;
				int c = edge[newp][j].cost;
				if (mark[t] == true)
					continue;
				if (dis[t] == -1 || dis[t] > dis[newp] + c)
					dis[t] = dis[newp] + c;
			}
			int min = 10000000;
			for (int j = 1; j <= n; j++)
			{
				if (mark[j] == true)
					continue;
				if (dis[j] == -1)
					continue;
				if (dis[j] < min)
				{
					min = dis[j]; newp = j;
				}
			}
			mark[newp] = true;
		}
		printf("%d\n", dis[n]);
	}
	return 0;
}

若是加上花費等條件，那麼只需要修改一下判定即可，例迪傑斯特拉算法

#include<stdio.h>
#include<vector>
using namespace std;
struct list
{
	int next;
	int distance;
	int cost;
};
vector<list> edge[101];
bool mark[101];
int dis[101];
int cost[101];
int main()
{
	int n, m;
	while (scanf("%d%d", &n, &m) != EOF)
	{
		if (m == 0 && n == 0)
			break;
		for (int i = 0; i <= n; i++)
			edge[i].clear();
		for (int i = 0; i < m; i++)
		{
			int a, b, c, d;
			scanf("%d%d%d%d", &a, &b, &c, &d);
			list temp;
			temp.distance = c;
			temp.cost = d;
			temp.next = a;
			edge[b].push_back(temp);
			temp.next = b;
			edge[a].push_back(temp);
		}
		for (int i = 0; i <= n; i++)
		{
			mark[i] = false;
			dis[i] = -1;
			cost[i] = 1000000;
		}
		int s, e;
		scanf("%d%d", &s, &e);
		dis[s] = 0;
		mark[s] = true;
		int newp = s;
		cost[s] = 0;
		for (int i = 0; i < n - 1; i++)
		{
			for (int j = 0; j < edge[newp].size(); j++)
			{
				int t = edge[newp][j].next;
				int c = edge[newp][j].distance;
				int co = edge[newp][j].cost;
				if (mark[t] == true)
					continue;
				if (dis[t] == -1 || dis[t] > dis[newp] + c || dis[t] == dis[newp] + c && cost[t] > cost[newp] + co)
				{
					dis[t] = dis[newp] + c;
					cost[t] = cost[newp] + co;
				}
			}
			int min = 10000000;
			for (int j = 1; j <= n; j++)
			{
				if (mark[j] == true)
					continue;
				if (dis[j] == -1)
					continue;
				if (dis[j] < min)
				{
					min = dis[j]; newp = j;
				}
			}
			mark[newp] = true;
		}
		printf("%d %d\n", dis[e], cost[e]);
	}
	return 0;
}