【NOI2017模拟4.4】最小直径

题目

Description

你有一个n个点m条边的森林，编号从0开始，边有边权，你现在要添加若干边权为L的边，满足：
1、最后n个点构成一颗树。
2、这棵树的直径尽量小。
请你求出这个最小的直径是多少。

Input

第一行三个整数n,m,L。
接下来m行，每行三个整数u,v,w，表示u和v之间有长为w的边。

Output

一行一个整数，表示最小直径的长度。

Sample Input

12 8 2
0 8 4
8 2 2
2 7 4
5 11 3
5 1 7
1 3 1
1 9 5
10 6 3

Sample Output

18

Data Constraint

对于20%的数据：n<=5；
对于60%的数据：n<=2000；
对于100%的数据：n<=50000,m

题解

比较感性而且显然的，我们可以先在每一颗树上找到一个离它最远的点最近的点，然后把每一颗树中的这样的点连起来变成一棵树，这棵树的直径就是最小的直径

贴代码

var
    a:array[0..150005,1..3]of longint;
    b:array[0..150005,1..2]of longint;
    ct,now:array[0..100005]of longint;
    cc:array[0..50005,1..4]of longint;
    bz,bq:array[0..50005]of boolean;
    h:array[0..200005]of longint;
    i,j,k,l,m,n,x,y,z,t1,t2,ans,ma1,ma2:longint;
Function Max(x,y:longint):longint;
begin
    if x>y then exit(x) else exit(y);
end;
procedure qsort(l,r:longint);
var
    i,j,mid:longint;
begin
    i:=l;
    j:=r;
    mid:=a[(i+j) div 2,1];
    repeat
        while a[i,1]<mid do inc(i);
        while a[j,1]>mid do dec(j);
        if i<=j then
        begin
            a[0]:=a[i]; a[i]:=a[j]; a[j]:=a[0];
            inc(i); dec(j);
        end;
    until i>j;
    if i<r then qsort(i,r);
    if l<j then qsort(l,j);
end;
procedure star;
begin
    b[a[1,1],1]:=1;
    for i:=2 to m do
    if a[i,1]<>a[i-1,1] then
    begin
        b[a[i-1,1],2]:=i-1;
        b[a[i,1],1]:=i;
    end;
    b[a[m,1],2]:=m;
end;
procedure dfs_size(x:longint);
var
    i,p:longint;
begin
    bz[x]:=true;
    for i:=b[x,1] to b[x,2] do
    if (bz[a[i,2]]=false) and (i<>0) then
    begin
        dfs_size(a[i,2]);
        p:=cc[a[i,2],1]+a[i,3];
        if p>cc[x,1] then
        begin
            cc[x,3]:=cc[x,1]; cc[x,4]:=cc[x,2];
            cc[x,1]:=p; cc[x,2]:=a[i,2];
        end else
        if p>cc[x,2] then
        begin
            cc[x,3]:=p;
            cc[x,4]:=a[i,2];
        end;
    end;
end;
procedure dfs_ag(x,z:longint);
var
    i,t,tt:longint;
begin
    bz[x]:=true;
    for i:=b[x,1] to b[x,2] do
    if (bz[a[i,2]]=false) and (i<>0) then
    begin
        if cc[x,2]=a[i,2] then t:=cc[x,3] else t:=cc[x,1];
        t:=max(t,z)+a[i,3];
        tt:=max(t,cc[a[i,2],1]);
        if tt<t1 then
        begin
            t1:=tt;
            t2:=a[i,2];
        end;
        dfs_ag(a[i,2],t);
    end;
end;
procedure bfs_go(x:longint);
var
    i,j,k,s1,s2:longint;
begin
    fillchar(bz,sizeof(bz),false);
    fillchar(ct,sizeof(ct),0);
    ans:=0;
    i:=1;
    j:=0;
    h[1]:=x;
    bz[x]:=true;
    while i>j do
    begin
        inc(j);
        s1:=h[j];
        for k:=b[s1,1] to b[s1,2] do
        if (bz[a[k,2]]=false) and (k<>0) then
        begin
            s2:=a[k,2];
            ct[s2]:=ct[s1]+a[k,3];
            if ct[s2]>ans then
            begin
                ans:=ct[s2];
                t1:=s2;
            end;
            bz[s2]:=true;
            inc(i);
            h[i]:=s2;
        end;
    end;
end;
begin
    assign(input,'diameter.in'); reset(input);
    assign(output,'diameter.out'); rewrite(output);
    readln(n,m,l);
    for i:=1 to m do
    begin
        readln(x,y,z);
        inc(x); inc(y);
        a[i,1]:=x; a[i,2]:=y; a[i,3]:=z;
        a[m+i,1]:=y; a[m+i,2]:=x; a[m+i,3]:=z;
    end;
    m:=m*2;
    qsort(1,m);
    star;
    for i:=1 to n do
    if (bz[i]=false) then dfs_size(i);
    fillchar(bz,sizeof(bz),false);
    for i:=1 to n do
    if (bz[i]=false) then
    begin
        t1:=cc[i,1]; t2:=i;
        dfs_ag(i,0);
        inc(now[0]);
        now[now[0]]:=t2;
    end;
    for i:=2 to now[0] do
    begin
        x:=m+i-1;
        a[x,1]:=now[1]; a[x,2]:=now[i]; a[x,3]:=l;
        x:=x+now[0]-1;
        a[x,1]:=now[i]; a[x,2]:=now[1]; a[x,3]:=l;
    end;
    m:=m+(now[0]-1)*2;
    qsort(1,m);
    star;
    bfs_go(1);
    bfs_go(t1);
    writeln(ans);
    close(input); close(output);
end.