传送门POJ3255

描述

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).

输入

Line 1: Two space-separated integers: N and RLines 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)

输出

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

样例

Input4 41 2 1002 4 2002 3 2503 4 100 Output450

题解

题意：输入n,m;1—n的点，m条边，找出次短路用dis[i][0]表示1到i的最短距离，dis[i][1]表示1到i的次短距离

Code

Dijkstra:

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960#include<bits/stdc++.h>using namespace std;#define LL long long#define INIT(a,b) memset(a,b,sizeof(a))const int N=5e3+7;const int R=1e5+7;const int inf=0x3fffffff;struct Edge{ int to,val,next;}edge[R<<1];struct Node{ int v,w; bool operator < (const Node &e)const{return w>e.w;}};int Begin[N],tot=0,vis[N];LL dis[N][2];int n,m;void add(int x,int y,int w){ edge[tot]=(Edge){y,w,Begin[x]}; Begin[x]=tot++;}void Dijkstra(int t){ priority_queue<Node> que; for(int i=1;i<=n;i++) dis[i][0]=dis[i][1]=inf; dis[t][0]=0; que.push((Node){t,dis[t][0]}); while(!que.empty()){ Node now=que.top();que.pop(); int x=now.v; if(dis[x][1]<now.w)continue; //剪枝 for(int i=Begin[x];~i;i=edge[i].next){ int ne=edge[i].to; int d=now.w+edge[i].val; //now.w不一定是到i的最短路 if(d<dis[ne][0]){ dis[ne][1]=dis[ne][0]; dis[ne][0]=d; que.push((Node){ne,dis[ne][0]}); } else if(d>dis[ne][0]&&d<dis[ne][1]){ dis[ne][1]=d; que.push((Node){ne,dis[ne][1]}); } } }}int main(){ ios::sync_with_stdio(false); cin.tie(0); INIT(Begin,-1);INIT(vis,0); tot=0; int x,y,w; cin>>n>>m; for(int i=0;i<m;i++){ cin>>x>>y>>w; add(x,y,w); add(y,x,w); } Dijkstra(1); cout<<dis[n][1]<<endl; return 0;}

SPFA:

12345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758#include<bits/stdc++.h>using namespace std;#define LL long long#define INIT(a,b) memset(a,b,sizeof(a))const int N=5e3+7;const int R=1e5+7;const int inf=0x3fffffff;struct Edge{ int to,val,next;}edge[R<<1];struct Node{ int v,w;};int Begin[N],tot=0;LL dis[N][2];int n,m;void add(int x,int y,int w){ edge[tot]=(Edge){y,w,Begin[x]}; Begin[x]=tot++;}void spfa(int t){ for(int i=1;i<=n;i++) dis[i][0]=dis[i][1]=inf; queue<Node> road; road.push((Node){t,0}); dis[t][0]=0;; while(!road.empty()){ Node top=road.front();road.pop(); int now=top.v; for(int i=Begin[now];~i;i=edge[i].next){ int ne=edge[i].to; LL d=top.w+edge[i].val; if(dis[ne][0]>d){ dis[ne][1]=dis[ne][0]; dis[ne][0]=d; road.push((Node){ne,dis[ne][0]}); } else if(dis[ne][0]<d && d<dis[ne][1]){ dis[ne][1]=d; road.push((Node){ne,dis[ne][1]}); } } }}int main(){ ios::sync_with_stdio(false); cin.tie(0); INIT(Begin,-1); int x,y,w; cin>>n>>m; for(int i=0;i<m;i++){ cin>>x>>y>>w; add(x,y,w); add(y,x,w); } spfa(1); cout<<dis[n][1]<<endl; return 0;}