POJ 2387 Til the Cows Come Home (图论,最短路径)
2017-07-22 20:54
567 查看
POJ 2387 Til the Cows Come Home (图论,最短路径)
Description
Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.
Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.
Input
Line 1: Two integers: T and NLines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.
Output
Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.Sample Input
5 51 2 20
2 3 30
3 4 20
4 5 20
1 5 100
Sample Output
90Http
POJ:https://vjudge.net/problem/POJ-2387Source
图论,最短路径解决思路
最短路径,直接用spfa可以解决代码
#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> #include<queue> #include<vector> using namespace std; const int maxN=1001; const int maxM=2001; const int inf=2147483647; class Edge { public: int v,w; }; int n,m; vector<Edge> E[maxN]; int Dist[maxN]; queue<int> Q; bool inqueue[maxN]; int main() { cin>>m>>n; for (int i=1;i<=m;i++) { int u,v,w; cin>>u>>v>>w; E[u].push_back((Edge){v,w}); E[v].push_back((Edge){u,w}); } for (int i=1;i<=n;i++) Dist[i]=inf; memset(inqueue,0,sizeof(inqueue)); Dist =0; inqueue =1; Q.push(n); do { int u=Q.front(); Q.pop(); inqueue[u]=0; for (int i=0;i<E[u].size();i++) { int v=E[u][i].v; int w=E[u][i].w; if (Dist[u]+w<Dist[v]) { Dist[v]=Dist[u]+w; if (inqueue[v]==0) { Q.push(v); inqueue[v]=1; } } } } while (!Q.empty()); cout<<Dist[1]<<endl; return 0; }
相关文章推荐
- POJ 2387 Til the Cows Come Home(最短路径)
- Poj 2387 Til the Cows Come Home(Dijkstra 最短路径)
- POJ 2387 Til the Cows Come Home --单源最短路径
- Poj 2387 Til the Cows Come Home(Dijkstra 最短路径)
- POJ 2387 Til the Cows Come Home Dijkstra求最短路径
- POJ 2387 Til the Cows Come Home -最短路径
- (阶段三 dijkstra算法温习 1.6)POJ 2387 Til the Cows Come Home(使用dijkstra算法求单源起点和单源终点的最短路径)
- POJ-2387 Til the Cows Come Home(Dijsktra算法求最短路径)
- poj 2387 Til the Cows Come Home(最短路径)
- poj2387——Til the Cows Come Home(最短路径)
- Til the Cows Come Home(poj 2387 Dijkstra算法(单源最短路径))
- POJ 2387 Til the Cows Come Home(最短路径,模板题)
- POJ_2387_Til the Cows Come Home(USACO 2004 November)_最短路径
- POJ 2387 Til the Cows Come Home (最短路径,Dijkstra算法)
- POJ 2387 Til the Cows Come Home(图论几个基本算法的初探)
- poj 2387 Til the Cows Come Home//最短路径
- [POJ](2387)Til the Cows Come Home---单源最短路径(图)
- POJ 2387 Til the Cows Come Home(Dijkstra算法)
- POJ 2387 Til the Cows Come Home (裸SPFA)
- Til the Cows Come Home POJ - 2387