Windows下用批次程式進行對拍

在做演算法類競賽的題目的時候，容易想到一個樸素的能保證完全正確的演算法，但是會逾時。而高效的演算法又不能保證完全寫對。這時候可以自己寫一個樸素的演算法、一個資料產生程式和一個檔案比較程式進行驗證高效演算法的正確性。

在Windows下，fc命令提供了比較檔案的功能，雖然批處理不如Linux下的bash等強大，但也足以寫個自動比較的程式了。

下面以求逆序對為例。題目類似於POJ3067，排序後求逆序對。資料量N=300000。容易想到類似於冒泡排序的枚舉逆序對，先寫出來。命名為standard.pas

const
    MAXN = 500000;
type 
    car = record x,v:longint end;

var
    a: array[0..MAXN] of car;
    n : longint;
    i, j : longint;
    sum : longint;

procedure swap(var a, b:car);
var
    temp:car;
begin
    temp := a;
    a := b;
    b := temp;
end;

begin
    readln(n);
for i:=1 to n do
        readln(a[i].x , a[i].v);
for i:=1 to n-1 do
for j:=i+1 to n do
if a[i].x > a[j].x then
                swap(a[i], a[j]);
for i:=1 to n-1 do
for j:=i+1 to n do
if a[i].v > a[j].v  then
                inc(sum);
    writeln(sum);
end.

用歸併排序等進階的演算法實現NLogN的程式，才能保證過掉這題。命名為題目要求的overtaking.pas。程式是用了歸併排序排好序，再把這個歸併排序複製一下改一下變成求逆序對的統計過程。

const
    MAXN = 500000;
type 
    car = record x,v:longint end;

var
    temp, a: array[0..MAXN] of car;
    n : longint;
    i : longint;
    sum : int64;

procedure swap(var a, b:car);
var
    temp:car;
begin
    temp := a;
    a := b;
    b := temp;
end;

procedure qsort(l, r:longint);
var
    i, j, m:longint;
begin
    i:=l;j:=r;m:=a[(l+r)div 2].x;
repeat
while (a[i].x < m) do inc(i);
while (a[j].x > m) do dec(j);
if i <= j then
begin
            swap(a[i],a[j]);
            inc(i);
            dec(j);
end;
until i > j;
if l < j then
        qsort(l, j);
if i < r then
        qsort(i, r);
end;

procedure merge(left, mid, right :longint);
var
    i, j, p, k:longint;
begin
    k := left;
    i := left;
    j := mid + 1;
while (i <= mid) and (j <= right) do
begin
if a[i].x <= a[j].x then
begin
            temp[k] := a[i];
            inc(i);
            inc(k);
end
else begin
            temp[k] := a[j];
            inc(j);
            inc(k);
end;
end;
for p := i to mid do
begin
        temp[k] := a[p];
        inc(k);
end;
for p := j to right do
begin
        temp[k] := a[p];
        inc(k);
end;
for i:= left to right do
        a[i] := temp[i];
end;

procedure merge_sort(left, right:longint);
var
    mid : longint;
begin
if left >= right then    
        exit;
    mid := (left + right) shr 1;
    merge_sort(left, mid);
    merge_sort(mid+1, right);
    merge(left, mid ,right);
end;



procedure merge2(left, mid, right :longint);
var
    i, j, p, k:longint;
begin
    k := left;
    i := left;
    j := mid + 1;
while (i <= mid) and (j <= right) do
begin
if a[i].v <= a[j].v then
begin
            temp[k] := a[i];
            inc(i);
            inc(k);
end
else begin
            inc(sum, mid - i + 1);
            temp[k] := a[j];
            inc(j);
            inc(k);
end;
end;
for p := i to mid do
begin
        temp[k] := a[p];
        inc(k);
end;
for p := j to right do
begin
        temp[k] := a[p];
        inc(k);
end;
for i:= left to right do
        a[i] := temp[i];
end;

procedure merge_sort2(left, right:longint);
var
    mid : longint;
begin
if left >= right then    
        exit;
    mid := (left + right) shr 1;
    merge_sort2(left, mid);
    merge_sort2(mid+1, right);
    merge2(left, mid ,right);
end;


procedure select_sort;
var 
    i, j:longint;
begin
for i:=1 to n-1 do
for j:=i+1 to n do
if a[i].x > a[j].x then
                swap(a[i], a[j]);
end;

begin

    readln(n);
for i:=1 to n do
begin
        readln(a[i].x, a[i].v);
end;
    merge_sort(1, n);
    merge_sort2(1, n);
    writeln(sum);


end.

然後寫一個資料產生器。此題造資料很簡單，產生隨機數就行了。

const
    INF = 199303170;
    MAXN = 3000;

var
    i, n:longint;
begin
    randomize;
    n := random(MAXN);
    writeln(n);
for i:=1 to n do
        writeln(random(INF),'',random(INF));
end.

然後寫一個批處理對拍，自動將這三個程式的輸入輸出從螢幕、鍵盤重新導向到檔案，然後比較standard.exe與overtaking產生的資料。

:loop
make.exe > data.txt
standard.exe < data.txt > std.txt
overtaking.exe < data.txt > ans.txt
fc std.txt ans.txt
if errorlevel 1 goto end
goto loop
:end

命令列介面會不斷提示找不到檔案差異，然後就可以放心提交了，當然前提是你樸素的演算法寫對了。