3408: [Usaco2009 Oct]Heat Wave 热浪

3408: [Usaco2009 Oct]Heat Wave 热浪

Time Limit: 3 Sec  Memory Limit: 128 MB
Submit: 67  Solved: 55
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Description

Input

 第1行：4个由空格隔开的整数T，C，Ts，Te.

 第2到第C+1行：第i+l行描述第i条道路．有3个由空格隔开的整数Rs，Re，Ci.

Output

    一个单独的整数表示Ts到Te的最小费用．数据保证至少存在一条道路．

Sample Input

7 11 5 4
2 4 2
1 4 3
7 2 2
3 4 3
5 7 5
7 3 3
6 1 1
6 3 4
2 4 3
5 6 3
7 2 1

Sample Output

7

HINT

Source

Gold

题解：话说USACO金组为啥时候冒出来这么多普及组题= =

最短路模板题不解释

 /**************************************************************
     Problem:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/

 type
     point=^node;
     node=record
                g,w:longint;
                next:point;
     end;
 var
    i,j,k,l,m,n,s,t,f,r:longint;
    a:array[..] of point;
    c,g:array[..] of longint;
    d:array[..] of longint;
    p:point;
 procedure add(x,y,z:longint);
           var p:point;
           begin
                new(p);p^.g:=y;p^.w:=z;
                p^.next:=a[x];a[x]:=p;
           end;
 begin
      readln(n,m,s,t);
      for i:= to n do a[i]:=nil;
      for i:= to m do
          begin
               readln(j,k,l);
               add(j,k,l);add(k,j,l);
          end;
      fillchar(g,sizeof(g),);
      fillchar(c,sizeof(c),);
      f:=;r:=;d[]:=s;c[s]:=;g[s]:=;
      while f<r do
            begin
                 p:=a[d[f]];
                 while p<>nil do
                       begin
                            if (c[p^.g]=) or ((c[p^.g]>) and (c[p^.g]>(c[d[f]]+p^.w))) then
                               begin
                                    c[p^.g]:=c[d[f]]+p^.w;
                                    if g[p^.g]= then
                                       begin
                                            g[p^.g]:=;
                                            d[r]:=p^.g;
                                            inc(r);
                                       end;
                               end;
                            p:=p^.next;
                       end;
                 g[d[f]]:=;
                 inc(f);
            end;
      for i:= to n do dec(c[i]);
      writeln(c[t]);
      readln;
 end.