POJ 3255 Roadblocks 次短路

题目链接:点击打开链接

题意:一个无向图,求从起点到终点的最短路(可以重复走),

思路:从分别以起点和原点做最短路,然后求枚举每条边,求出比最短路长的最短的哪一条.

cpp:点击打开链接

#include <cmath>#include <iostream>#include <map>#include <vector>#include <cstring>#include <queue>#include <cstdio>#include <algorithm>using namespace std;const int INF=1e9;const int maxn=10010;struct Edge{int from,to,dist;Edge(int u,int v,int d):from(u),to(v),dist(d) {}};struct HeapNode{int d,u;bool operator < (const HeapNode& rhs) const{return d > rhs.d;}};struct Dijkstra{int n,m;vector<Edge> edges;vector<int> G[maxn];bool done[maxn];int d[maxn];void init(int n){this->n=n;for(int i=0;i<n;i++) G[i].clear();edges.clear();}void addEdges(int from,int to,int dist){edges.push_back(Edge(from,to,dist));m=edges.size();G[from].push_back(m-1);}void dijkstra(int s){priority_queue<HeapNode> Q;for(int i=0;i<n;i++) d[i]=INF;d[s]=0;memset(done,0,sizeof(done));HeapNode tep;tep.d=0,tep.u=s;Q.push(tep);while (!Q.empty()){HeapNode x=Q.top();Q.pop();int u=x.u;if(done[u]) continue;done[u]=true;for(int i=0;i<G[u].size();i++){Edge& e=edges[G[u][i]];if(d[e.to]>d[u]+e.dist){d[e.to]=d[u]+e.dist;tep.d=d[e.to],tep.u=e.to;Q.push(tep);}}}}};Dijkstra dij1,dij2;int main (){int ans;int tp,a,b,n,m;int temp;//freopen("data.in","r",stdin);    while (~scanf("%d%d",&n,&m))    {dij1.init(n);dij2.init(n);        for (int i=1;i<=m;i++)        {            scanf("%d%d%d",&a,&b,&tp);dij1.addEdges(a-1,b-1,tp);dij2.addEdges(b-1,a-1,tp);dij2.addEdges(a-1,b-1,tp);dij1.addEdges(b-1,a-1,tp);        }dij1.dijkstra(0);dij2.dijkstra(n-1);        ans=INF;for(int i=0;i<dij1.edges.size();i++){Edge tem=dij1.edges[i];temp=dij1.d[tem.from]+dij2.d[tem.to]+tem.dist;if(temp==dij1.d[n-1]){temp+=2*tem.dist;}ans=min(ans,temp);}cout<<ans<<endl;    }return 0;}