Roadblocks ———POJ - 3255 （次短路）

題目：

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)

 

題意：

給出一些點和它們之間的距離，求從1號點到n號點次短路的距離

 

思路：

1到n的次短路長度必然產生於：從1走到x的最短路+e[x][y]+y到n的最短路
利用spfa算法先求出1到每一個節點的最短路，和n到每一個節點的最短路
然後枚舉每一條邊作爲中間邊（x，y）或者（y，x），如果加起來長度等於最短路長度則跳過，否則更新
從1走到x的最短路+e[x][y]+y到n的最短路  給dis1[n] 比較 找大於dis1[n] 且是最小的那一個 

 

代碼：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define ll long long
#define N 200861
#define inf 0x3f3f3f3f
using namespace std;
int n,m,dis1[N],dis2[N],first[N],k,vis[N];
struct node
{
    int u,v,w,next;
}e[N];
void Inint()
{
    k=1;
    memset(dis1,inf,sizeof(dis1));
    memset(dis2,inf,sizeof(dis2));
    memset(first,-1,sizeof(first));
}
void add(int u,int v,int w)//鄰接表存圖
{
    e[k].u=u;
    e[k].v=v;
    e[k].w=w;
    e[k].next=first[u];
    first[u]=k++;
}
void spfa(int u,int dis[])
{
    memset(vis,0,sizeof(vis));
    queue<int>q;
    dis[u]=0;
    vis[u]=1;
    q.push(u);
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=first[u];i!=-1;i=e[i].next)
        {
            int v=e[i].v;
            int w=e[i].w;
            if(dis[v]>dis[u]+w)
            {
                dis[v]=dis[u]+w;
                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
}
int main()
{
    Inint();
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        add(u,v,w);
        add(v,u,w);
    }
    spfa(1,dis1);//求1到n最短路
    spfa(n,dis2);//求n到1最短路
    int minn=inf;
    for(int i=1;i<=n;i++)
    {
        for(int j=first[i];j!=-1;j=e[j].next)
        {
            int v=e[j].v;
            int w=e[j].w;
            int t=dis1[i]+dis2[v]+w;
            if(t>dis1[n]&&t<minn)
                minn=t;
        }
    }
    printf("%d\n",minn);
    return 0;
}