莫队算法讲解

　　莫队算法的大体思路就是暴力的转移，尽量的减少转移的时间。

　　假设我们求出了区间[l1,r1]的答案，那么对于区间[l1,r1+1]我们可以o(1)的转移，对于不同的询问，我们将l当做横坐标，r当做纵坐标，这样建立的一张图，求最小manhattan生成树，需要转移的时间是最少的。

　　但是求一颗manhattan最小生成树的时间已经比较长了，所以我们退而求其次。将n个询问分成floor(sqrt(n))个块，对于每个块我们按照r排序，然后每个块之间暴力的转移，经莫涛证明，时间复杂度是O(n^1.5)的。理论上所有的区间询问都可以用这种方法尝试。

　　经典的应用为bzoj 2038：http://61.187.179.132/JudgeOnline/problem.php?id=2038

/**************************************************************

    Problem: 2038

    User: BLADEVIL

    Language: Pascal

    Result: Accepted

    Time:3136 ms

    Memory:2572 kb

****************************************************************/

 

//By BLADEVIL

type

    rec                         =record

        l, r, w, s              :longint;

    end;

     

var

 

    n, m                        :longint;

    c, size                     :array[0..50010] of int64;

    len                         :longint;

    a                           :array[0..50010] of rec;

    now                         :longint;

    col, ans                    :array[0..50010] of int64;

    all, num                    :int64;

     

procedure swap(var a,b:longint);

var

    c                           :longint;

begin

    c:=a; a:=b; b:=c;

end;

 

procedure swap_rec(var a,b:rec);

var

    c                           :rec;

begin

    c:=a; a:=b; b:=c;

end;

 

function gcd(a,b:int64):int64;

begin

    if a<b then exit(gcd(b,a)) else

    if b=0 then exit(a) else exit(gcd(b,a mod b));

end;

 

procedure qs(low,high:longint);

var

    i, j, xx, yy                :longint;

begin

    i:=low; j:=high; xx:=a[(i+j) div 2].w;

    yy:=a[(i+j) div 2].r;

    while i<j do

    begin

        while (a[i].w<xx) or (a[i].w=xx) and (a[i].r<yy) do inc(i);

        while (a[j].w>xx) or (a[j].w=xx) and (a[j].r>yy) do dec(j);

        if i<=j then

        begin

            swap_rec(a[i],a[j]);

            inc(i); dec(j);

        end;

    end;

    if i<high then qs(i,high);

    if j>low then qs(low,j);

end;

     

procedure init;

var

    i                           :longint;

     

begin

    read(n,m);

    for i:=1 to n do read(c[i]);

    len:=trunc(sqrt(m));

    for i:=1 to m do

    begin

        read(a[i].l,a[i].r);

        if a[i].l>a[i].r then swap(a[i].l,a[i].r);

        size[i]:=a[i].r-a[i].l+1;

        a[i].w:=a[i].l div len+1;

        a[i].s:=i;

    end;

    qs(1,m);

end;

     

procedure main;

var

    i, j                        :longint;

begin

    i:=1;

    while i<=m do

    begin

        now:=a[i].w;

        fillchar(col,sizeof(col),0);

        for j:=a[i].l to a[i].r do

        begin

            ans[a[i].s]:=ans[a[i].s]+2*(col[c[j]]);

            col[c[j]]:=col[c[j]]+1;

        end;

        inc(i);

        while a[i].w=now do

        begin

            ans[a[i].s]:=ans[a[i-1].s];

            for j:=a[i-1].r+1 to a[i].r do

            begin

                ans[a[i].s]:=ans[a[i].s]+2*(col[c[j]]);

                col[c[j]]:=col[c[j]]+1;

            end;

            if a[i-1].l<a[i].l then

            begin

                for j:=a[i-1].l to a[i].l-1 do

                begin

                    col[c[j]]:=col[c[j]]-1;

                    ans[a[i].s]:=ans[a[i].s]-2*col[c[j]];

                end;

            end else

                for j:=a[i].l to a[i-1].l-1 do

                begin

                    ans[a[i].s]:=ans[a[i].s]+2*(col[c[j]]);

                    col[c[j]]:=col[c[j]]+1;

                end;

            inc(i);

            if i>m then break;

        end;

    end;

    for i:=1 to m do

    begin

        if size[i]=1 then all:=1 else all:=size[i]*(size[i]-1);

        num:=gcd(ans[i],all);

        writeln(ans[i] div num,'/',all div num);

    end;

end;

     

     

begin

    init;

    main;

end.