poj3255基础次短路

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Author      :Crazy_AC(JamesQi)
Time        :
File Name   :
思路：分别以源点1和终点N为源点，两次SPFA求得dist1[i](1到各点的最短距离)以及dist2[i](各点到N的最短距离),
然后就是枚举边了，设某一条边的两端为u,v,权值为w,则dist1[u]+w+dist2[v]即为1->N的一条路径，在所有的路径中
找出次短的即可。
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#include <iostream>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
using namespace std;
#define FILL(a,b) memset(a,b,sizeof a)
#define CLR(a) memset(a,0,sizeof a)
template<class T> inline T Get_Max(const T&a,const T&b) {return a < b?b:a;}
template<class T> inline T Get_Min(const T&a,const T&b) {return a < b?a:b;}
const int inf = 1 << 30;
const int maxn = 5000;
struct Edge{
	int to,w;
	Edge(){}
	Edge(int to,int w):to(to),w(w){}
};
int dis1[maxn],dis2[maxn];
int mark[maxn];
int n,m;
vector<vector<Edge> > G;
void spfa(int st,int dist[])
{
	memset(mark,0,(n + 2)*sizeof(int));
	fill(dist,dist + n + 2,inf);
	dist[st] = 0;
	queue<int> que;
	que.push(st);
	mark[st] = 1;
	while(!que.empty())
	{
		int u = que.front();
		mark[u] = 0;
		que.pop();
		for (int i = 0;i < G[u].size();i++)
		{
			int v = G[u][i].to;
			int w = G[u][i].w;
			if (dist[v] > dist[u] + w)
			{
				dist[v] = dist[u] + w;
				if (!mark[v])
				{
					mark[v] = 1;
					que.push(v);
				}
			}
		}
	}
}
int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);
	while(scanf("%d%d",&n,&m) != EOF)
	{
		G.clear();
		G.resize(n + 2);
		int u,v,w;
		while(m--)
		{
			scanf("%d%d%d",&u,&v,&w);
			G[u].push_back(Edge(v,w));
			G[v].push_back(Edge(u,w));
		}
		spfa(1,dis1);
		spfa(n,dis2);
		int ans1 = inf,ans2 = inf;
		for (int u = 1;u <= n;u++)
		{
			for (int i = 0;i < G[u].size();i++)
			{
				int v = G[u][i].to;
				int w = G[u][i].w;
				if (dis1[u] + dis2[v] + w < ans1)
				{
					ans2 = ans1;
					ans1 = dis1[u] + dis2[v] + w;
				}
				else if (dis1[u] + dis2[v] + w < ans2 && dis1[u] + dis2[v] + w != ans1)
				{
					ans2 = dis1[u] + dis2[v] + w;
				}
			}
		}
		printf("%d\n",ans2);
	}
	return 0;
}