Evanyou Blog 彩带

　　题目传送门

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)

---

　　分析：

　　一句话题意，次短路模板。

　　一般来说做次短路可以通过先跑一遍最短路，然后记录路径，在最短路路径上每次删去其中一条边，然后再跑一遍最短路求出次短路。但是有一种更加简洁快速的方法，用$Dijkstra$一次求出最短路与次短路。

　　思路非常简单，在求最短路的同时开另外一个数组记录次短路，每次被更新的最短路就可以更新到次短路里面，或者是更新的路径比最短路长但比当前记录的次短路要短时也要更新。不过还有一些细节要注意，具体可以看代码。

　　Code：

//It is made by HolseLee on 17th Aug 2018
//POJ3255
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<queue>
#include<algorithm>
#define Max(a,b) (a)>(b)?(a):(b)
#define Min(a,b) (a)<(b)?(a):(b)
#define Swap(a,b) (a)^=(b)^=(a)^=(b)
using namespace std;

const int N=5005;
const int M=1e5+7;
typedef pair<int,int> P;
int n,m,head[N],siz,dis[N],dist[N];
struct Node{
    int to,val,nxt;
}edge[M<<1];
priority_queue<P,vector<P>,greater<P> > T;

inline int read()
{
    char ch=getchar();int num=0;bool flag=false;
    while(ch<'0'||ch>'9'){if(ch=='-')flag=true;ch=getchar();}
    while(ch>='0'&&ch<='9'){num=num*10+ch-'0';ch=getchar();}
    return flag?-num:num;
}

inline void add(int x,int y,int z)
{
    edge[++siz].to=y;
    edge[siz].val=z;
    edge[siz].nxt=head[x];
    head[x]=siz;
}

void dijkstra()
{
    memset(dis,0x7f,sizeof(dis));
    memset(dist,0x7f,sizeof(dist));
    dis[1]=0;
    T.push(P(1,0));
    int x,y,d,dt;
    while(!T.empty()){
        x=T.top().first,d=T.top().second;T.pop();
        if(dist[x]<d)continue;
        for(int i=head[x];i!=-1;i=edge[i].nxt){
            y=edge[i].to;
            dt=d+edge[i].val;
            if(dis[y]>dt){
                Swap(dis[y],dt);
                T.push(P(y,dis[y]));
            }
            if(dist[y]>dt&&dis[y]<dt){
                dist[y]=dt;
                T.push(P(y,dist[y]));
            }
        }
    }
}

int main()
{
    n=read();m=read();
    int x,y,z;
    memset(head,-1,sizeof(head));
    for(int i=1;i<=m;++i){
        x=read(),y=read(),z=read();
        add(x,y,z);add(y,x,z);
    }
    dijkstra();
    printf("%d
",dist[n]);
    return 0;
}