題面

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 N

• Lines 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 5
  1 2 20
  2 3 30
  3 4 20
  4 5 20
  1 5 100

  Sample Output

  90

題解

題目大意：給定N個點，T條邊
求出從節點1到節點N的最短路徑長度。

---

直接求最短路即可
習慣用SPFA。。。
如果用dijkstra要考慮重邊的情況（舍掉之類的）

#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<queue> #include<algorithm> using namespace std; #define MAX 11000 #define MAXL 22000 struct Line {int v,next,w; }e[MAXL]; int u,v,w; int h[MAX],cnt=1; int T,N; queue<int> Q;bool vis[MAX]; int dis[MAX]; inline void Add(int u,int v,int w) {e[cnt]=(Line){v,h[u],w};h[u]=cnt++; } int main() {cin>>T>>N;for(int i=1;i<=T;++i){cin>>u>>v>>w;Add(u,v,w);Add(v,u,w);}for(int i=1;i<=N;++i)dis[i]=1050000000;/*********SPFA***********/vis[1]=true;dis[1]=0;Q.push(1);while(!Q.empty()){u=Q.front();Q.pop();vis[u]=false;for(int i=h[u];i;i=e[i].next){v=e[i].v;if(dis[v]>dis[u]+e[i].w){dis[v]=dis[u]+e[i].w;if(!vis[v]){vis[v]=true;Q.push(v);}}}}cout<<dis[N]<<endl;return 0; }

轉載于:https://www.cnblogs.com/cjyyb/p/7237109.html

總結

以上是生活随笔為你收集整理的【POJ2387】Til the Cows Come Home （最短路）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。