POJ 3255：Roadblocks（次短路）

题目大意：求无向图的次短路。

分析：

  在起点终点各求一次最短路，枚举边，通过该边的最短路为其权值加上到起点和终点最短路之和，找到最短但又比最短路长的路径。

代码：

program block;
type
  point=^node;
    node=record
    v,c:longint; next:point; end;
var
  a:array[0..5000]of point;
  dis:array[0..5000,1..2]of longint;
  q:array[0..5000]of longint;
  g:array[0..5000]of boolean;
  e:array[0..100000,1..3]of longint;
  n,i,m,x,y,v,ans:longint;  p:point;
procedure add(x,y,c:longint);
var p:point;
begin
  new(p); p^.v:=y; p^.c:=c; p^.next:=a[x];  a[x]:=p;
end;
procedure spfa(s,d,l:longint);
var i,h,t,u,v,k:longint;
begin
  fillchar(g,sizeof(g),false);
  for i:=1 to n do dis[i,l]:=maxlongint div 3;  dis[s,l]:=0; g[s]:=true;
  h:=0; t:=1; q[1]:=s;
  while h<t do
   begin
     inc(h); u:=q[h]; g[u]:=false; new(p); p:=a[u];
     while p<>nil do
     begin
       v:=p^.v;
       if dis[v,l]>dis[u,l]+p^.c then
        begin
          dis[v,l]:=dis[u,l]+p^.c;
          if g[v]=false then begin g[v]:=true; inc(t);q[t]:=v;end;
        end;
       p:=p^.next;
     end;
   end;
end;
begin
  assign(input,'block.in');
  reset(input);
  assign(output,'block.out');
  rewrite(output);
  readln(n,m);
  for i:=1 to m do
   begin
     readln(x,y,v);  e[i,1]:=x; e[i,2]:=y; e[i,3]:=v;
     add(x,y,v); add(y,x,v);
   end;
  spfa(1,n,1); spfa(n,1,2); ans:=maxlongint;
  for i:=1 to m do
   begin
    v:=dis[e[i,1],1]+dis[e[i,2],2]+e[i,3];   if (v<ans)and(v>dis[n,1]) then ans:=v;
    v:=dis[e[i,2],1]+dis[e[i,1],2]+e[i,3];   if (v<ans)and(v>dis[n,1]) then ans:=v;
   end;
  writeln(ans);
  close(input); close(output);
end.

 

来源：https://www.cnblogs.com/qtyytq/p/5328597.html