POJ 3255 Roadblocks 无向图次短路

Roadblocks

Description

Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

Input

Line 1: Two space-separated integers: N and R 
Lines 2..R+1: Each line contains three space-separated integers: A, B, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000)

Output

Line 1: The length of the second shortest path between node 1 and node N

Sample Input

4 4

1 2 100

2 4 200

2 3 250

3 4 100

Sample Output

450

Hint

Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)

Source

​​USACO 2006 November Gold​​

算法分析：

题意：

n个点 m条无向边 从点1走到点n 求出无向图次短路径的长度

有向图次短路：​​点击打开链接​​

分析：

1.用数组d存储点1到其他点的最短路径长度

2.用数组rd存储点n到其他点的最短路径长度

3.枚举每条边，两个短点分别到起点终点的最短长度加上这条边的长度 记录次短路径的值

代码分析：

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>  
#include<list>  
using namespace std;  
const double eps = 1e-8;  
typedef long long LL;  
typedef unsigned long long ULL;  
const int INF = 0x3f3f3f3f;  
const int INT_M_INF = 0x7f7f7f7f;  
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;  
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;  
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};  
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};  
const int MOD = 1e9 + 7;  
const double pi = acos(-1.0);  

const int MAXN=5010;  
const int MAXM=100010;

using namespace std;

struct Edge
{
  int v,w,next;
}edge[MAXM*2];
int d[MAXN],rd[MAXN];
int tot,head[MAXN];
int vis[MAXN];
int n,m;
 
void init()           //初始化
{
    tot=0;
    memset(head,-1,sizeof(head));
    for(int i=0;i<=n;i++)
        d[i]=rd[i]=INF;
}
 
void addedge(int u,int v,int w)
{
    edge[tot].v=v;
    edge[tot].w=w;
    edge[tot].next=head[u];
    head[u]=tot++;
}
 
void SPFA(int s,int dis[])
{
    int u,v;
    queue<int> q;
    memset(vis,0,sizeof(vis));
    dis[s]=0; vis[s]=1;
    q.push(s);
    while(!q.empty())
    {
        u=q.front(); q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            v=edge[i].v;
            if(dis[u]+edge[i].w<dis[v])
            {
                dis[v]=dis[u]+edge[i].w;
                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        while(m--)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);        //无向图
            addedge(v,u,w);
        }
        SPFA(1,d);    
        SPFA(n,rd);    
        int ans=INF;
        for(int i=1;i<=n;i++)
        {
            for(int j=head[i];j!=-1;j=edge[j].next)
            {
                int v=edge[j].v;
                int w=edge[j].w;
                int  tmp=d[i]+rd[v]+w;
                if(tmp>d[n]&&ans>tmp)
                    ans=tmp;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}