POJ3255 (secondary short circuit)

Topic Link: Click to open the link

Problem Solving Ideas:

According to the Dijkstra thought to do the second short circuit, the first time with the adjacency table, note that the two-way side of the problem and the subscript of the node to 1.

Full code:

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include < cmath> #include <set> #include <vector> #include <climits> #include <queue>using namespace Std;typedef Long Long ll;const int maxn = 100001;typedef pair<int, int> p;struct edge{int to, cost;};  int INF = 99999999;int N, r;vector<edge> g[maxn];int dist[maxn];int 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 p = que.top ();        Que.pop ();        int v = p.second, d = p.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) {swap (dist[e.to], D2);               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 () {#ifdef Doubleq freopen ("In.txt", "R", stdin);        #endif while (CIN >> n >> r) {int A, B, D;            for (int i = 0; i < R; i + +) {cin >> a >> b >> D;            A--;            B--;            struct Edge temp;            Temp.to = b;            Temp.cost = D;            G[a].push_back (temp);            struct Edge temp2;            Temp2.to = A;            Temp2.cost = D;        G[b].push_back (TEMP2);    } solve (); } return 0;}

More highlights please visit: Click to open the link

POJ3255 (secondary short circuit)