本文的基础知识见：算法导论第24章：单源最短路径

入门题：HDOJ 2544 最短路

Dijkstra算法

算法实现：邻接矩阵 + 维护结点的d值 + 每次取min{d}入队 & 松弛

运行时间：未用优先队列，O(V²)

#include <iostream>
#include <cstring>
using namespace std;

const int inf = 1e8;
int edge[105][105];
int d[105];
bool vis[105];
int n,m,a,b,c,cur,imin;

void init()
{
	memset(edge,0,sizeof(edge));
	memset(vis,0,sizeof(vis));
	for(int i=1; i<105; ++i)
		d[i] = inf;
	cur = 1;
	vis[1] = 1;
	d[1] = 0;

	for(int i=0; i<m; ++i)
	{
		cin>>a>>b>>c;
		if(edge[a][b]==0 || edge[a][b] > c)
			edge[a][b] = edge[b][a] = c;
	}
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	while(cin>>n>>m && n)
	{
		init();

		for(int i=1; i<=n-1; ++i)
		{
			for(int j=1; j<=n; ++j)
			{
				if(vis[j] == 0 && edge[cur][j] &&  d[j] > d[cur] + edge[cur][j])
					d[j] = d[cur] + edge[cur][j];
			}

			imin = inf;
			for(int i=1; i<=n; ++i)
			{
				if(vis[i]==0 && imin > d[i])
				{
					imin = d[i];
					cur = i;
				}
			}

			if(cur == n) break;
			vis[cur] = 1;
		}

		cout<<d[n]<<"\n";
	}

	return 0;
}

Bellman-Ford算法

Edge结构体 + 松弛所有边V-1次

#include <iostream>
using namespace std;

struct Edge
{
	int begin,end,length;
} edge[10005];

int n,m,a,b,c;
const int inf = 1e8;
int d[105];

void init()
{
	for(int i=1; i<105; ++i)
		d[i] = inf;

	for(int i=0; i<m; ++i)
		cin>>edge[i].begin>>edge[i].end>>edge[i].length;

	d[1] = 0;
}

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);

	while(cin>>n>>m && n)
	{
		init();

		for(int i=1; i<=n-1; ++i)
		{
			for(int j=0; j<m; ++j)
			{
				if(d[edge[j].begin] < inf && d[edge[j].begin] + edge[j].length < d[edge[j].end])
					d[edge[j].end] = d[edge[j].begin] + edge[j].length;
				if(d[edge[j].end] < inf && d[edge[j].end] + edge[j].length < d[edge[j].begin])
					d[edge[j].begin] = d[edge[j].end] + edge[j].length;
			}
		}

		cout<<d[n]<<"\n";
	}

	return 0;
}