Bzoj 1774 [Usaco2009 Dec]Toll 过路费

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原题网址：http://www.lydsy.com/JudgeOnline/problem.php?id=1774
题目简述：求两点之间最短路，特别的，最短路定义为边权和加最大点权。(n≤250)
题解：看数据范围，可以考虑Floyd，一般Floyd最外层循环枚举的是最大标号的中转点，在这题中，不妨以点权从小到大的顺序枚举，在中转点点权比两端点点权大的时候，说明中转点是路径上权值最大的点，这时可以更新答案。

const  INF=maxlongint div 3;var  dis,ans:array[0..251,0..251] of longint;  w,num:array[0..251] of longint;  n,m,q,i,j,k,a,b,l:longint;procedure swap(var a,b:longint);  var t:longint;  begin t:=a;a:=b;b:=t;end;procedure sort(l,r:longint);  var    i,j,e:longint;  begin    i:=l;j:=r;e:=w[(l+r)>>1];    repeat      while w[i]<e do inc(i);      while e<w[j] do dec(j);      if not (i>j) then        begin          swap(w[i],w[j]);          swap(num[i],num[j]);          inc(i);dec(j);        end;    until i>j;    if l<j then sort(l,j);    if i<r then sort(i,r);  end;function min(a,b:longint):longint;  begin if (a<b) then exit(a) else exit(b); end;begin  read(n,m,q);  for i:=1 to n do    for j:=1 to m do      dis[i][j]:=INF;  for i:=1 to n do    dis[i][i]:=0;  ans:=dis;  for i:=1 to n do read(w[i]);  for i:=1 to n do num[i]:=i;  sort(1,n);  for i:=1 to m do    begin      read(a,b,l);      dis[a][b]:=min(dis[a][b],l);      dis[b][a]:=min(dis[b][a],l);    end;  for k:=1 to n do    for i:=1 to n do      for j:=1 to n do        begin          if (w[k]>=w[i])and(w[k]>=w[j])            then ans[num[i]][num[j]]:=min(ans[num[i]][num[j]],dis[num[i]][num[k]]+dis[num[k]][num[j]]+w[k]);          dis[num[i]][num[j]]:=min(dis[num[i]][num[k]]+dis[num[k]][num[j]],dis[num[i]][num[j]]);        end;  for i:=1 to q do    begin      read(a,b);      writeln(ans[a][b]);    end;end.

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