POJ-3255 roadblocks

To solve the problem of short-circuit, the method is similar to finding the shortest single source, but the subject is the single source shortest circuit and the second shortest piece solution

To a certain point of the shortest path (EG:U): Assuming that the most short-circuit is s->v->u, the short-circuit is s->v->u ' or s->v '->u Note: S->v ' indicates a secondary short-circuit of s->v, v->u ' means v->u Secondary short Circuit

What needs to be done is to update the single source shortest and second shortest

If the current node is U, the node adjacent to U is V

1, if the node with u adjacent to the V, through u can minimize the distance of S to its, then update its shortest, and note the original S to v shortest circuit, in case of an updated short-circuit

2, a, if the path length of s through U to V is greater than the minimum path length of S to V, but less than S to V of the secondary short-circuit length, then update the second short path of S to V

b, if the original shortest circuit less than the short-circuit, update the secondary short-circuit

#include <iostream> #include <vector> #include <queue> #define MAXN 100010#define INF 9999999999using    Namespace Std;typedef pair<int,int>p;struct Edge {int from;    int to; Long long cost;}; int N,r;vector <edge> G[maxn];long long dist[maxn],dist2[maxn];void Solve () {priority_queue<p,vector<p    >,greater<p> > que;    Fill (Dist,dist+n,inf);    Fill (Dist2,dist2+n,inf);    Dist[0] = 0;    Que.push (P (0,0));        while (!que.empty ()) {p pp = que.top ();        Que.pop ();        int v = pp.second;        int d = Pp.first;        if (Dist2[v] < D) continue;            for (int i=0;i<g[v].size (); i++) {Edge &e = G[v][i];            int d2 = d+ e.cost;                if (Dist[e.to] > D2) {int TP = dist[e.to];                Dist[e.to] = D2;                D2 = TP;            Que.push (P (dist[e.to],e.to));              } if (Dist2[e.to] > D2 && dist[e.to] <d2) {  DIST2[E.TO]=D2;            Que.push (P (dist2[e.to],e.to)); }}} cout<< Dist2[n-1]<<endl;}    int main () {cin>>n>>r;    Edge TMP,TMP1;    int from,to,cost;        for (int i=0;i<r;i++) {cin>>from>>to>>cost;        cout<<tmp.from<<tmp.to<<tmp.cost;        Tmp.from = from-1;        tmp.to = to-1;        Tmp.cost =cost;        G[from-1].push_back (TMP);        Tmp1.from = to-1;        tmp1.to = from-1;        Tmp1.cost = Cost;    G[to-1].push_back (TMP1);    } solve (); return 0;}

　　

POJ-3255 roadblocks