和Dijksta求最短路一樣，只是要維護兩個數組：最短路d1,次短路d2。然后更新的時候注意細節。

//#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 #elsefreopen("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; } View Code

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轉載于:https://www.cnblogs.com/BMan/p/3647537.html

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