对拍程序

对拍程序就是比较两个程序的输出是否一致

拿P1144 最短路计数当例子

先写个构造数据的程序 

如

#include <stdio.h>
#include<cstring>
#include<ctime>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#define mx ((int)1e9)
using namespace std;

char name[30],cmd[30];
int a[66666];
void pp(int n)
{
    printf("%d
",rand()%(n/3)+n/3*2);
}
int main()
{
    srand(time(0));
    for(int i=1;i<=1;i++)
    {
        //sprintf(name,"data%d.in",i);
        //freopen(name,"w",stdout);
        cout << "100 1000" <<endl;
        for(int j=1;j<=1000;j++)
        {
            int x,y;
            if(j<100)
            {
                x=j+1;
                y=rand()%(j)+1;
            }
            else
            {
                x=rand()%100+1;
                y=rand()%100+1;
            }
            printf("%d %d
",x,y);
        }
 
        //fclose(stdout);
        //sprintf(cmd,"./number <data%d.in >data%d.out",i,i);
        //system(cmd);
    }
}

再写两个解问题的代码

一个暴力一个正解

正解

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<cmath>
#include<set>
#include<map>
#include<complex>
#define T 111111
#define M 100003
using namespace std;
int sc()
{
    int i=0; char c=getchar();
    while( c>'9' || c<'0' ) c=getchar();
    while( c>='0'&& c<='9') i=i*10+c-'0',c=getchar();
    return i;
}
int f[T],dis[T],mark[T],q[T];
vector<int>e[T];
int n,m,l,r;
int main()
{
    //freopen("map.in","r",stdin);
    //freopen("map.out","w",stdout);
    n=sc(); m=sc();
    for(int i=1; i<=m; i++)
    {
        int x=sc(),y=sc();
        e[x].push_back(y);
        e[y].push_back(x);
    }
    q[1]=1; l=1; r=2;
    memset(dis,11,sizeof(dis));
    mark[1]=dis[1]=f[1]=1;
    while( l<r )
    {
        int k=q[l++];
        for(int i=0; i<e[k].size(); i++)
            if( dis[e[k][i]]>dis[k] )
            {
                dis[e[k][i]]=dis[k]+1;
                f[e[k][i]]+=f[k];
                f[e[k][i]]%=M;
                if(!mark[e[k][i]])
                {
                    mark[e[k][i]]=1;
                    q[r++]=e[k][i];
                }
            }
    }
    for(int i=1; i<=n; i++) printf("%d
",f[i]%M);
    return 0;
}

暴力

/** @file map.cpp
 *  This is an implementation file for a precompiled header.
 */

/* LANG can be C++11 or C++14 for those more recent releases */

// 17.4.1.2 Headers

// C

#ifndef _GLIBCXX_NO_ASSERT
#include <cassert>
#endif

#include <cctype>
#include <cerrno>
#include <cfloat>
#include <ciso646>
#include <climits>
#include <clocale>
#include <cmath>
#include <csetjmp>
#include <csignal>
#include <cstdarg>
#include <cstddef>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>

#if __cplusplus >= 201103L
#include <ccomplex>
#include <cfenv>
#include <cinttypes>
#include <cstdalign>
#include <cstdbool>
#include <cstdint>
#include <ctgmath>
#include <cwchar>
#include <cwctype>
#endif

// C++

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <iterator>
#include <limits>
#include <list>
#include <locale>
#include <map>
#include <memory>
#include <new>
#include <numeric>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <typeinfo>
#include <utility>
#include <valarray>
#include <vector>

#if __cplusplus >= 201103L
#include <array>
#include <atomic>
#include <chrono>
#include <condition_variable>
#include <forward_list>
#include <future>
#include <initializer_list>
#include <mutex>
#include <random>
#include <ratio>
#include <regex>
#include <scoped_allocator>
#include <system_error>
#include <thread>
#include <tuple>
#include <typeindex>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#endif
using namespace std;
const int maxn=1e5+50,maxm=2e5+50,modp=100003;
int n,m,head[maxn],cnt,vis[maxn],dis[maxn];
struct edge{
	int fr,ne,to;
}e[maxm*2];
void add(int u,int v)
{
	e[++cnt].fr=u;
	e[cnt].ne=head[u];
	e[cnt].to=v;
	head[u]=cnt;
}
queue<int>q;
void bfs()
{
	q.push(1);
	vis[1]=1;
	dis[1]=0;
	int u,fr,to;
	while(!q.empty())
	{
		u=q.front();q.pop();
		for(int i=head[u];i;i=e[i].ne)
		{
			fr=e[i].fr;to=e[i].to;
			if((to!=1&&!dis[to]))
			{
				dis[to]=dis[fr]+1;
				q.push(to);
			}
			if(dis[fr]+1==dis[to])
			{
				vis[to]+=vis[fr];
				vis[to]%=modp;
			}
		}
	}
}
int main()
{
	//freopen("map.in","r",stdin);
	//freopen("map.out","w",stdout);
	scanf("%d%d",&n,&m);
	for(int i=1,u,v;i<=m;i++)
	{
		scanf("%d%d",&u,&v);
		add(u,v);add(v,u);
	}
	bfs();
	for(int i=1;i<=n;i++)
	{
		printf("%d
",vis[i]);
	}
}

对拍程序

#include<windows.h>
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
	int t=100;
	while(t--)
	{
		system("a > a.in");//a是生成数据的程序//a.in读入文件
		system("map < a.in > 1.out");map是正解//out两个输出文件
		system("1   < a.in > 2.out");1是暴力
		if(system("fc 1.out 2.out")) break;
	}
	system("pause");
	return 0;
}

最后生成4个exe放到同个文件夹运行对拍的exe