和Dijksta求最短路一样,只是要维护两个数组:最短路d1,次短路d2。然后更新的时候注意细节。
//#pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<sstream> #include<cmath> #include<climits> #include<string> #include<map> #include<queue> #include<vector> #include<stack> #include<set> using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; #define pb(a) push(a) #define INF 0x1f1f1f1f #define lson idx<<1,l,mid #define rson idx<<1|1,mid+1,r #define PI 3.1415926535898 template<class T> T min(const T& a,const T& b,const T& c) { return min(min(a,b),min(a,c)); } template<class T> T max(const T& a,const T& b,const T& c) { return max(max(a,b),max(a,c)); } void debug() { #ifdef ONLINE_JUDGE #else freopen("in.txt","r",stdin); //freopen("d:\\out1.txt","w",stdout); #endif } int getch() { int ch; while((ch=getchar())!=EOF) { if(ch!=‘ ‘&&ch!=‘\n‘)return ch; } return EOF; } struct HeapNode { int d,u; bool operator < (const HeapNode &ant) const { return ant.d<d; } }; struct Edge { int from,to,dist; }; const int maxn=5005; vector<int> g[maxn]; vector<Edge> edge; int n; int d1[maxn],d2[maxn]; void init() { for(int i=1;i<=n;i++) g[i].clear(); edge.clear(); } void add(int u,int v,int w) { Edge e=(Edge){u,v,w}; edge.push_back(e); g[u].push_back(edge.size()-1); } void solve(int s) { for(int i=1;i<=n;i++) d1[i]=d2[i]=INF; priority_queue<HeapNode> q; d1[s]=0; q.push((HeapNode){0,s}); while(!q.empty()) { HeapNode x=q.top(); q.pop(); if(x.d>d2[x.u])continue; int u=x.u; for(int i=0;i<g[u].size();i++) { Edge &e=edge[g[u][i]]; int v=e.to; int d=x.d; if(e.dist+d<d1[v]) { d2[v]=d1[v]; d1[v]=d+e.dist; q.push((HeapNode){d1[v],v}); }else if(e.dist+d<d2[v]&&e.dist+d!=d1[v]) { d2[v]=d+e.dist; q.push((HeapNode){d2[v],v}); } } } } int main() { int m; while(scanf("%d%d",&n,&m)!=EOF) { init(); 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); } solve(1); printf("%d\n",d2[n]); } return 0; }