POJ2253

题意:给出一个无向图,求一条1~2的路径使得路径上的最大边权最小.

分析:dijkstra变形,将更新距离的过程改为取最大值即可.

code:

const maxn=201;
var   d:array[0..maxn,0..maxn] of extended;
      x,y:array[0..maxn] of longint;
      v:array[0..maxn] of boolean;
      n,s,t,i,j,p,casenum:longint;
      minx:extended;

      function calc(a,b:longint):extended;
      begin
            calc:=sqrt(sqr(x[a]-x[b])+sqr(y[a]-y[b]));
      end;

      function max(a,b:extended):extended;
      begin
            if a>b then exit(a); exit(b);
      end;

      function min(a,b:extended):extended;
      begin
            if a>b then exit(b); exit(a);
      end;

begin
      while not seekeof do
      begin
            readln(n);
            if n=0 then break;
            inc(casenum);
            for i:=1 to n do readln(x[i],y[i]);
            readln;
            for i:=1 to n-1 do
               for j:=i+1 to n do
               begin
                     d[i,j]:=calc(i,j);
                     d[j,i]:=d[i,j];
               end;

            fillchar(v,sizeof(v),0);
            s:=1;
            t:=2;
            v[s]:=true;
            for j:=2 to n do
            begin
                  minx:=1e10;
                  for i:=1 to n do
                     if (not v[i])and(d[s,i]<>0)and(d[s,i]<minx) then
                     begin
                           minx:=d[s,i];
                           p:=i;
                     end;
                  v[p]:=true;
                  for i:=1 to n do
                     if (d[s,p]<>0)and(d[p,i]<>0) then
                       d[s,i]:=min(d[s,i],max(d[s,p],d[p,i]));
            end;
            writeln('Scenario #',casenum);
            writeln('Frog Distance = ',d[1,2]:0:3);
            writeln;
      end;
end.