POJ3255:Roadblocks(次短路 SPFA+A星)

时间：2014-05-01 03:34:24 收藏：0 阅读：533

给出1～N 个点的距离，求从1号到N号的次短路，直接用k短路来做了，，dj会TLE，用spfa就过了

题目：

I - Roadblocks

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Description

Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

Input

Line 1: Two space-separated integers: N and R
Lines 2..R+1: Each line contains three space-separated integers: A, B, and D that describe a road that connects intersections A and B and has lengthD (1 ≤ D ≤ 5000)

Output

Line 1: The length of the second shortest path between node 1 and node N

Sample Input

Sample Output

Hint

Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)

代码：

  1 #include <iostream>
  2 #include <algorithm>
  3 #include <vector>
  4 #include <queue>
  5 using namespace std;
  6 
  7 #define MAXN 5000+10
  8 
  9 
 10 int N,R;
 11 
 12 #define MP(a,b) make_pair(a,b)
 13 #define INF 100000000
 14 
 15 #define pii pair<int,int>
 16 vector<pii> G[MAXN];
 17 struct pri
 18 { 
 19     bool operator () (const pair<int,int> &p1,const pair<int,int>&p2)
 20     {
 21         return p1.second< p2.second;
 22     }
 23 };
 24 
 25 int d[MAXN];
 26 void dj(int s)
 27 {
 28     for(int i = 1;i<=N;i++) 
 29     {
 30         d[i]= INF;
 31         d[i] = INF;
 32     }
 33     d[s]=0;
 34     d[s]=0;
 35     
 36     priority_queue<pii,vector<pii>,pri> q;
 37     q.push( MP(s,d[s]) );
 38     while(!q.empty())
 39     {   
 40         int u = q.top().first;
 41         q.pop();
 42 
 43         int size = G[u].size();
 44 
 45         for(int i=0;i<size;i++)
 46         {
 47             int v,w;
 48             v = G[u][i].first;
 49             w = G[u][i].second;
 50 
 51             if( d[v] > d[u]+w )
 52             {
 53                 d[v]=d[u]+w;
 54                 q.push(MP(v,d[v]));
 55 
 56             }
 57         }
 58     }
 59 
 60 }
 61 /*
 62 void spfa(int src)
 63 {
 64     
 65     for(int i=1;i<=N;i++)
 66     {
 67         d[i] = INF;
 68     }
 69     d[src]=0;
 70     queue<int> q;
 71     q.push(src);
 72     int a,next;
 73     int vis[MAXN]={0};
 74     while(!q.empty())
 75     {
 76         a = q.front();
 77         q.pop();
 78         vis[a] = 0;
 79         int len = G[a].size();
 80         for(int i=0;i<len;i++)
 81         {
 82             next = G[a][i].first;
 83             if( d[next]> d[a]+G[a][i].second )
 84             {
 85                 d[next] = d[a]+G[a][i].second;
 86                 if( !vis[next] )
 87                 {
 88                     q.push(next);
 89                     vis[next]=1;
 90                 }
 91             }
 92         }
 93     }
 94 }
 95 */
 96 struct node{
 97 
 98     int now,g,h,f;
 99     bool operator < (const node& x) const
100     {
101         if( f == x.f )
102         {
103             return x.g<g;
104         }
105         else
106         {
107             return x.f<f;
108         }
109     }
110 };
111 void astar(int src,int to)
112 {
113     priority_queue<node> q;
114     node a, next;
115 
116     a.now = src;
117     a.g = 0;
118     a.f = d[src]+a.g;
119     q.push(a);
120     int cnt = 0;
121     while(!q.empty())
122     {
123         a = q.top();
124         q.pop();
125         if( a.now == to )
126         {
127             cnt++;
128             if( cnt == 2 )
129             {
130                 cout<<a.g<<endl;
131                 return;
132             }
133         }
134         int size = G[a.now].size();
135         for(int i=0;i<size;i++)
136         {
137             next.now = G[a.now][i].first;
138             next.g = a.g + G[a.now][i].second;
139             next.f = next.g+d[next.now];
140             q.push(next);
141         }
142     }
143 }
144 
145 int main()
146 {
147 
148     cin>>N>>R;
149 
150     for(int i=0;i<R;i++)
151     {
152         int f,t,w;
153         cin>>f>>t>>w;
154 
155         G[f].push_back( MP(t,w) );
156         G[t].push_back( MP(f,w) );
157     }
158     
159     dj(N);    
160 //    spfa(N);
161     astar(1,N);
162     return 0;
163 }

POJ3255:Roadblocks(次短路 SPFA+A星),码迷,mamicode.com