[最短路]pku2387-Til the Cows Come Home

http://acm.pku.edu.cn/JudgeOnline/problem?id=2387

 

 

Til the Cows Come Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8663		Accepted: 2864

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 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

Hint

INPUT DETAILS:

There are five landmarks.

OUTPUT DETAILS:

Bessie can get home by following trails 4, 3, 2, and 1.

Source

USACO 2004 November

 

 

题目大意：给出一个无向图（可能有重边），求点n到点1的最短路径。

分析：简单题。不过，本题有很猥琐的地方：先输入边数，再输入点数.不注意的话就会re了。

 

codes:

 

var tot,n,m,k:longint; e:array[1..4000] of record b,w:longint; end; d,f,ne:array[0..4000] of longint; b:array[0..5000] of longint; v:array[0..1000] of boolean; procedure init; var i,x,y,w:longint; begin tot:=0; readln(m,n); for i:=1 to m do begin readln(x,y,w); inc(tot); ne[tot]:=f[x]; f[x]:=tot; e[tot].w:=w; e[tot].b:=y; inc(tot); ne[tot]:=f[y]; f[y]:=tot; e[tot].w:=w; e[tot].b:=x; end; end; procedure spfa; var head,tail,i,tx:longint; begin head:=0; tail:=1; fillchar(v,sizeof(v),false); fillchar(d,sizeof(d),127); b[1]:=1; v[1]:=true; d[1]:=0; repeat inc(head); tx:=b[head]; i:=f[tx]; while i<>0 do begin if d[tx]+e[i].w<d[e[i].b] then begin d[e[i].b]:=d[tx]+e[i].w; if not v[e[i].b] then begin inc(tail); b[tail]:=e[i].b; end; end; i:=ne[i]; end; v[tx]:=false; until head>=tail; end; procedure main; var i:longint; ans:int64; begin ans:=0; spfa; writeln(d[n]); end; begin init; main; end.