部分经典算法

双关键字快排

procedure swap(a,b:longint);
var t:longint;
begin
for i:=1 to 3 do
begin
t:=x[a,i];
x[a,i]:=x[b,i];
x[b,i]:=t;
end;
end;

procedure sort(l,r: longint);
var i,j,x1,x2,y: longint;
begin
i:=l; j:=r; x1:=x[(l+r) div 2,1]; x2:=x[(l+r) div 2,2];
repeat
while (x[i,1]<x1)or((x[i,1]=x1)and(x[i,2]<x2)) do inc(i);
while (x1<x[j,1])or((x[j,1]=x1)and(x[j,2]>x2)) do dec(j);
if not(i>j) then
begin
swap(i,j);
inc(i); dec(j);
end;
until i>j;
if l<j then sort(l,j);
if i<r then sort(i,r);
end;

归并排序

逆序对

var
x,y:array[0..2000000]of longint;
i,j,k:longint;
n,len:longint;
ans:int64;
procedure sort(l,r:longint);
var i,j,mid:longint;
begin
if l>=r then exit;
mid:=(l+r)>>1;
sort(l,mid);
sort(mid+1,r);
i:=l; j:=mid+1; len:=l-1;
while (i<=mid)or(j<=r) do
if (j>r)or((i<=mid)and(x[i]<=x[j]))
then begin inc(len); y[len]:=x[i]; inc(i); end
else begin inc(len); y[len]:=x[j]; inc(j); inc(ans,mid-i+1);  {inc是用来算逆序对的} end;
for i:=l to r do
x[i]:=y[i];
end;

begin
readln(N);
for i:=1 to n do
read(x[i]);
ans:=0;
sort(1,n);
writeln(ans);
end.

树状数组

var
x,y:array[0..1000005]of longint;
i,j,k:longint;
n,maxn:longint;
ans:int64;
function query(a:longint):int64;
var tt:int64;
begin
tt:=0;
while a>0 do
begin
inc(tt,y[a]);
dec(a,a and(-a));
end;
exit(tt);
end;

procedure update(a:longint);
begin
while a<=maxn do
begin
inc(y[a]);
inc(a,a and(-a));
end;
end;

begin
readln(n); maxn:=0;
for i:=1 to n do
begin read(x[i]); if x[i]>maxn then maxn:=x[i]; end;
ans:=0;
for i:=n downto 1 do
begin
inc(ans,query(x[i]-1));
update(x[i]);
end;
writeln(ans);
end.

堆排序

var
n,i:longint;
x,heap:array[0..1000000]of longint;
procedure shai(z:longint);
var tt,a,b,s:longint;
begin
tt:=z;
while (heap[tt*2]<>-maxlongint)or(heap[tt*2+1]<>-maxlongint) do
begin
a:=tt*2; b:=a+1;
if (heap[a]>heap[tt])and(heap[a]>=heap[b]) then begin s:=heap[a]; heap[a]:=heap[tt]; heap[tt]:=s; tt:=a; end else
if (heap[b]>heap[tt])and(heap[b]>=heap[a]) then begin s:=heap[b]; heap[b]:=heap[tt]; heap[tt]:=s; tt:=b; end
else break;
end;
end;

procedure prepare;
begin
readln(n);
for i:=1 to 5*n do
heap[i]:=-maxlongint;
for i:=1 to n do
read(heap[i]);
end;

begin
prepare;
heap[0]:=n;
for i:=(n div 2) downto 1 do
shai(i);
while heap[0]<>1 do
begin
inc(x[0]); x[x[0]]:=heap[1];
dec(heap[0]); heap[1]:=0; shai(1);
end;
x
:=heap[1];
for i:=n downto 2 do
write(x[i],' ');
writeln(x[1]);
end.