POJ3255 Roadblocks(次短路)

 

POJ3255 Roadblocks(次短路)

分类： 最短路径 2013-10-08 17:22  91人阅读  评论(0)  收藏  举报

http://poj.org/problem?id=3255

题意:某街区共有R条道路、N个路口。道路可以双向通行。问1号路口到N号路口的次短路长度是多少? 次短路指的是比最短路长度长的次短的路径。同一条边可以经过多次。

 #include <cstdio>
 #include <cstring>
 #include <vector>
 #include <queue>
 #include <algorithm>
 using namespace std;

 const int maxn = 5000 + 10;
 const int INF = 1000000000;
 typedef pair<int, int> P;
 int N, R;
 struct edge {
     int to, cost;
 };
 vector<edge> G[maxn];
 int dist[maxn];  ///save minimum path
 int dist2[maxn];  //次短路经
 void solve() {
     priority_queue<P, vector<P>, greater<P> > que;
     fill(dist, dist+N, INF);  //memset
     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));
             }
         }
     }
     printf("%d\n",dist2[N-1]);
 }

 int main() {
     int i, a, b, c;
     edge e;
     while(~scanf("%d%d",&N,&R)) {
         for(i=0; i<R; ++i) {
             scanf("%d%d%d", &a, &b, &c);
             a--,b--;
             e.to = b, e.cost = c;
             G[a].push_back(e);
             e.to = a;
             G[b].push_back(e);
         }
         solve();
     }
     return 0;
 }