windows和Ubuntu下的对拍和测试多组样例

C++

使用system

在linux下，终端shell语言

#include<bits/stdc++.h>
using namespace std;

int main()
{
	system("sudo g++ 1.cpp -o my");
	system("sudo g++ 2.cpp -o std");
	system("sudo g++ 3.cpp -o ran");
	while (1){
		system("./ran > data");
		system("./my < data > my.out");
		system("./std < data > std.out");
		if (system("diff std.out my.out")){
			cout << "WA" << endl;
			break;
		}
		else cout << "AC" << endl;
	}
	return 0;
}

shell语言

【Shell脚本攻略】-shell中各种括号()、(())、[]、[[]]、{}的作用

Shell 输入/输出重定向

Shell $0, $#, $*, $@, $?, $$和命令行参数

shell变量和引用

循环选择

测试运行时间

#!/bin/bash

g++ random.cpp -o random
g++ my.cpp -o my
g++ std.cpp -o std

i=1

while true; do
	i=$((i+1))
    ./random > in${i}.in
    ./std < in${i}.in > std.out
    ./my  < in${i}.in > my.out
    diff std.out my.out
    if [ $? -ne 0 ]
    then 
        echo test ${i} WA
        break
    else
        echo test ${i} AC
    fi
done

测运行时间我是拒绝的 <- time ./a 就可以测速度了。。。

upd.2019.10.29 感谢 ymz 的提醒。

Bat

还是BATCH好玩

@echo off
g++ std.cpp -o std
g++ my.cpp -o my
g++ ran.cpp -o ran

set /a i=0

:loop
set /a i=%i%+1
ran > tst.txt
my < tst.txt > my.out
std < tst.txt > std.out
fc /w my.out std.out
if not errorlevel 1 goto aa
goto bb
:aa
echo test %i% AC
goto loop
:bb
echo test %i% WA
pause

测试运行时间

测时间还是在c++内部搞搞好。

#include<cstdio>
#include<ctime>
#include<windows.h>
// #include<unistd.h> in linux
using namespace std;

int main()
{
	clock_t st, ed;
	double cost;
	st = clock();
	// main
	ed = clock();
	cost = ((double)(ed-st))/CLOCKS_PER_SEC;
	printf("%lf", cost);
	return 0;
}

SPJ

@echo off
g++ my.cpp -o my.exe
g++ spj.cpp -o spj.exe


set i=-1

:loop
set /a i=%i%+1
echo %i% > bala.out
call my.exe < bala.out > my.out
call spj.exe < my.out > spj.out
fc /w spj.out bala.out
if not errorlevel 1 (
echo test %i% AC
goto loop
)
echo test %i% WA
pause

上面是一道例题的spj：构造一个迷宫使(0, 0) -> (n-1, m-1) 的最短路长度是k。(k <= 1000, n, m <= 50)

构造程序：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N = 100;
int k;
int n, m, rem, mp[N][N];

void Initialize(int row)
{
	memset(mp, 0, sizeof(mp));
	for (int i = 1; i <= 49; i += 2)
		for (int j = 1; j <= row; j++)
			mp[j][i] = 1;
	for (int i = 2; i <= 49; i += 2)
		mp[(i>>1&1 ? row : 1)][i] = 1;
}

int main()
{
	cin >> k;
	if (k <= 49){
		cout << "1 " << k+1 << endl;
		for (int i = 1; i <= k+1; i++) cout << '.'; cout << endl;
		return 0;
	}
	if (k <= 98){
		int nn = (k+2)/2, mm = k+2-nn;
		cout << nn << " " << mm << endl;
		for (int i = 1; i <= mm; i++) cout << '.'; cout << endl;
		for (int i = 2; i <= nn; i++){
			for (int j = 1; j < mm; j++)
				cout << '#';
			cout << '.' << endl;
		}
		return 0;
	}
	if (k&1^1){ // k even
		n = 49;
		m = 49;
		rem = 49*49-24*48-k;
		Initialize(49);
		int i = 2;
		while (rem){
			if (48*2 <= rem){
				mp[1][i] = 1;
				rem -= 48*2;
				i += 4;
			}
			else{
				rem >>= 1;
				mp[49-rem][i] = 1;
				rem = 0;
			}
		}
	}
	else{ // k odd
		n = 50;
		m = 49;
		rem = 50*49-24*49-k;
		Initialize(50);
		int i = 2;
		while (rem){
			if (49*2 <= rem){
				mp[1][i] = 1;
				rem -= 49*2;
				i += 4;
			}
			else{
				rem >>= 1;
				mp[50-rem][i] = 1;
				rem = 0;
			}
		}
	}
	cout << n << " " << m << endl;
	for (int i = 1; i <= n; i++){
		for (int j = 1; j <= m; j++)
			cout << (mp[i][j] ? '.' : '#');
		cout << endl;
	}
	return 0;
}

spj：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
const int N = 100;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
int n, m, mp[N][N], dis[N][N];
queue<pair<int, int> > q;

int main()
{
	cin >> n >> m;
	memset(mp, 0, sizeof(mp));
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++){
			char c = getchar();
			while (c != '.' && c != '#') c = getchar();
			mp[i][j] = (c == '.');
		}
	q.push(make_pair(1, 1));
	dis[1][1] = 0;
	while (!q.empty()){
		int x = q.front().first;
		int y = q.front().second;
		q.pop();
		for (int i = 0; i < 4; i++){
			int nx = x+dx[i];
			int ny = y+dy[i];
			if (mp[nx][ny]){
				q.push(make_pair(nx, ny));
				dis[nx][ny] = dis[x][y]+1;
				mp[nx][ny] = 0;
			}
		}
	}
	cout << dis[n][m] << endl;
	return 0;
}

很快就把所有的输入数据全都跑完了，稳~