一开始用的是先按怨气值由大到小排序，再判断二分图的方法，但是超时，多次调试未果，只好上网找了题解，下面介绍一个博客上的方法：http://blog.sina.com.cn/s/blog_918b1f91010175i2.html
program prison;
type
node=record
x,y,v:longint;
end;
var
n,m,i,fx,fy:longint;
a:array[1..100000] of node;
father:array[1..40000] of longint;
procedure quick(left,right:longint);
var
i,j,temp:longint;
k:node;
begin
i:=left;
j:=right;
temp:=a[(i+j) div 2].v;
while i<=j do
begin
while a[i].v>temp do inc(i);
while a[j].v<temp do dec(j);
if i<=j then
begin
k:=a[i];
a[i]:=a[j];
a[j]:=k;
inc(i);
dec(j);
end;
end;
if left<j then quick(left,j);
if i<right then quick(i,right);
end;
function findfather(k:longint):longint;
begin
if father[k]=k then findfather:=k
else
begin
father[k]:=findfather(father[k]);
findfather:=father[k];
end;
end;
begin
assign(input,'prison.in');
reset(input);
assign(output,'prison.out');
rewrite(output);
fillchar(a,sizeof(a),0);
fillchar(father,sizeof(father),0);
readln(n,m);
for i:=1 to m do
readln(a[i].x,a[i].y,a[i].v);
quick(1,m);
for i:=1 to n*2 do
father[i]:=i;
for i:=1 to m do
begin
fx:=findfather(a[i].x);
fy:=findfather(a[i].y);
if fx=fy then
begin
writeln(a[i].v);
close(input);
close(output);
halt;
end;
father[fy]:=findfather(a[i].x+n);
father[fx]:=findfather(a[i].y+n);
end;
writeln('0');
close(input);
close(output);
end.
调试的时候发现如果把father[i]的初始值设为-1，把函数findfather中的if father[k]=k改为if father[k]=-1的话，会有4个点栈溢出，其原因未明，希望各位大牛指教
