Roadblocks（次短路经）

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)

第一次接触次短路的问题，原来接触过次小生成树的问题，现在也忘了。
求解次短路也是按着最短路的套路来。
从1到n的次短路等于dis1[i]+p[i][j]+dis2[j],其中dis1代表着顶点一到顶点i的最短路，dis2代表着n到j的最短路，再加上之间的距离就好了。其中用的vector，spfa求解最短路
代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;const int maxx=5e3+10;
struct node{int v;int w;
}pp;
vector<node>p[maxx];
int dis1[maxx];
int dis2[maxx];
int vis[maxx];
int n,m;void spfa(int u,int *dis)
{queue<int >q;memset(vis,0,sizeof(vis));vis[u]=1;dis[u]=0;q.push(u);while(!q.empty()){u=q.front();q.pop();for(int i=0;i<p[u].size();i++){int v=p[u][i].v;int w=p[u][i].w;if(dis[v]>dis[u]+w){dis[v]=dis[u]+w;if(!vis[v]){vis[v]=1;q.push(v);}}}}
}int main()
{scanf("%d%d",&n,&m);memset(dis1,inf,sizeof(dis1));memset(dis2,inf,sizeof(dis2));for(int i=0;i<m;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);pp.v=b;pp.w=c;p[a].push_back(pp);pp.v=a;pp.w=c;p[b].push_back(pp);}spfa(1,dis1);spfa(n,dis2);int ans=inf;for(int i=1;i<=n;i++){for(int j=0;j<p[i].size();j++){int v=p[i][j].v;int w=p[i][j].w;int ant=dis1[i]+dis2[v]+w;if(ant<ans&&ant>dis1[n]){ans=ant;}}}printf("%d\n",ans);
}

努力加油a啊，(^o)/~

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