Java教程

对拍 详解

本文主要是介绍对拍 详解,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

对拍详解

对拍是算法竞赛中除了瞪眼法以外 最有效的 d e b u g debug debug 利器
它可以通过随机数据大体验证算法的正确性
比测大水样例有效多了


这里我拿洛谷P3628 [APIO2010]特别行动队(一道斜优板子)作为例题
要想对拍,首先你需要准备:

  1. 一个写好的暴力(一定要保证正确)
  2. 一个写好待测的正解
  3. 一个datamaker
  4. 一个checker

暴力:

写出暴力是对拍的必要条件
对于例题我们需要先写一个暴力 d p dp dp
代码:

#include<bits/stdc++.h>
#define ll long long
#define int ll
using namespace std;
const int maxn=1e6+5;
int n,dp[maxn],a,b,c,sum[maxn];
inline ll read()
{
	ll ret=0;char ch=' ',c=getchar();
	while(!(c<='9'&&c>='0')) ch=c,c=getchar();
	while(c<='9'&&c>='0') ret=(ret<<1)+(ret<<3)+c-'0',c=getchar();
	return ch=='-'?-ret:ret;
}
signed main()
{
	n=read();a=read();b=read();c=read();
	for(int i=1;i<=n;i++) sum[i]=sum[i-1]+read();
	for(int i=1;i<=n;i++) dp[i]=-2e18;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=i;j++) dp[i]=max(dp[i],dp[j]+a*(sum[i]-sum[j])*(sum[i]-sum[j])+b*(sum[i]-sum[j])+c);
	printf("%lld",dp[n]);
	return 0;	
}

并起名为 d p . c p p dp.cpp dp.cpp

正解

然后我们写出斜优正解
代码:

#include<bits/stdc++.h>
#define ll long long
#define int ll
using namespace std;
const int maxn=1e6+5;
int n,dp[maxn],a,b,c,sum[maxn],q[maxn];
inline ll read()
{
	ll ret=0;char ch=' ',c=getchar();
	while(!(c<='9'&&c>='0')) ch=c,c=getchar();
	while(c<='9'&&c>='0') ret=(ret<<1)+(ret<<3)+c-'0',c=getchar();
	return ch=='-'?-ret:ret;
}
int X(int i) {return 2*a*sum[i];}
int Y(int i) {return dp[i]+a*sum[i]*sum[i]-b*sum[i]+c;}
int deltaX(int i,int j) {return X(i)-X(j);}
int deltaY(int i,int j) {return Y(i)-Y(j);}
signed main()
{
	n=read();a=read();b=read();c=read();
	for(int i=1;i<=n;i++) sum[i]=sum[i-1]+read();
	for(int i=1;i<=n;i++) dp[i]=-2e18;
	int head=0,tail=0;
	for(int i=1;i<=n;i++)
	{
		while(head<tail&&deltaY(q[head+1],q[head])>=sum[i]*deltaX(q[head+1],q[head])) head++;
		dp[i]=dp[q[head]]+a*(sum[i]-sum[q[head]])*(sum[i]-sum[q[head]])+b*(sum[i]-sum[q[head]])+c;
		while(head<tail&&deltaY(q[tail],q[tail-1])*deltaX(i,q[tail])>=deltaY(i,q[tail])*deltaX(q[tail],q[tail-1])) tail--;
		q[++tail]=i;
	}
	printf("%lld",dp[n]);
	return 0;	
}

并命名为 s t d . c p p std.cpp std.cpp

datamaker

然后问我们就要写datamaker了
我们通过 r a n d rand rand 函数(具体用法请自行百度)来造随机数据
代码:

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,a,b,c;
inline ll read()
{
	ll ret=0;char ch=' ',c=getchar();
	while(!(c<='9'&&c>='0')) ch=c,c=getchar();
	while(c<='9'&&c>='0') ret=(ret<<1)+(ret<<3)+c-'0',c=getchar();
	return ch=='-'?-ret:ret;
}
int main()
{
	srand(time(0));
	n=rand()%10000;
	a=rand()%4-5;
	b=rand()%20000000-10000000;
	c=rand()%20000000-10000000;
	printf("%d\n%d %d %d\n",n,a,b,c);
	while(n--) printf("%d ",rand()%100+1);
	return 0;
}

并命名为 d a t a m a k e r . c p p datamaker.cpp datamaker.cpp

checker

最关键的就是 c h e c k e r checker checker
它起到类似评测机的作用

#include<bits/stdc++.h>
#define ll long long
using namespace std;
inline ll read()
{
	ll ret=0;char ch=' ',c=getchar();
	while(!(c<='9'&&c>='0')) ch=c,c=getchar();
	while(c<='9'&&c>='0') ret=(ret<<1)+(ret<<3)+c-'0',c=getchar();
	return ch=='-'?-ret:ret;
}
int main()
{
	for(int i=1;i<=10000;i++)
	{
		system("datamaker.exe>data.txt");
		system("dp.exe<data.txt>dp.txt");
		double st=clock();
		system("std.exe<data.txt>std.txt");
		double ed=clock();
		if(system("fc dp.txt std.txt")) {printf("WA\n");break;}
		else printf("AC #%d Time:%.3lfms\n",i,ed-st);
	}
	return 0;
}

然后我们运行 c h e c k e r checker checker 即可完成对拍
注意:我们每次改完正解后要重新编译!!!
在这里插入图片描述
这是在 w i n d o w s windows windows 环境下
在 l i n u x linux linux 环境下也类似
dp:

#include<bits/stdc++.h>
#define ll long long
#define int ll
using namespace std;
const int maxn=1e6+5;
int n,dp[maxn],a,b,c,sum[maxn];
inline ll read()
{
    ll ret=0;char ch=' ',c=getchar();
    while(!(c<='9'&&c>='0')) ch=c,c=getchar();
    while(c<='9'&&c>='0') ret=(ret<<1)+(ret<<3)+c-'0',c=getchar();
    return ch=='-'?-ret:ret;
}
signed main()
{
    freopen("data.in","r",stdin);
    freopen("test.out","w",stdout);
    n=read();a=read();b=read();c=read();
    for(int i=1;i<=n;i++) sum[i]=sum[i-1]+read();
    for(int i=1;i<=n;i++) dp[i]=-2e18;
    for(int i=1;i<=n;i++)
	for(int j=0;j<=i;j++) dp[i]=max(dp[i],dp[j]+a*(sum[i]-sum[j])*(sum[i]-sum[j])+b*(sum[i]-sum[j])+c);
    printf("%lld",dp[n]);
    return 0;	
}

std:

#include<bits/stdc++.h>
#define ll long long
#define int ll
using namespace std;
const int maxn=1e6+5;
int n,dp[maxn],a,b,c,sum[maxn],q[maxn];
inline ll read()
{
    ll ret=0;char ch=' ',c=getchar();
    while(!(c<='9'&&c>='0')) ch=c,c=getchar();
    while(c<='9'&&c>='0') ret=(ret<<1)+(ret<<3)+c-'0',c=getchar();
    return ch=='-'?-ret:ret;
}
int X(int i) {return 2*a*sum[i];}
int Y(int i) {return dp[i]+a*sum[i]*sum[i]-b*sum[i]+c;}
int deltaX(int i,int j) {return X(i)-X(j);}
int deltaY(int i,int j) {return Y(i)-Y(j);}
signed main()
{
    freopen("data.in","r",stdin);
    freopen("std.out","w",stdout);
    n=read();a=read();b=read();c=read();
    for(int i=1;i<=n;i++) sum[i]=sum[i-1]+read();
    for(int i=1;i<=n;i++) dp[i]=-2e18;
    int head=0,tail=0;
    for(int i=1;i<=n;i++)
    {
	while(head<tail&&deltaY(q[head+1],q[head])>=sum[i]*deltaX(q[head+1],q[head])) head++;
	dp[i]=dp[q[head]]+a*(sum[i]-sum[q[head]])*(sum[i]-sum[q[head]])+b*(sum[i]-sum[q[head]])+c;
	while(head<tail&&deltaY(q[tail],q[tail-1])*deltaX(i,q[tail])>=deltaY(i,q[tail])*deltaX(q[tail],q[tail-1])) tail--;
	q[++tail]=i;
    }
    printf("%lld",dp[n]);
    return 0;	
}

datamaker:

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,a,b,c;
inline ll read()
{
  
    ll ret=0;char ch=' ',c=getchar();
    while(!(c<='9'&&c>='0')) ch=c,c=getchar();
    while(c<='9'&&c>='0') ret=(ret<<1)+(ret<<3)+c-'0',c=getchar();
    return ch=='-'?-ret:ret;
}
int main()
{
    freopen("data.in","w",stdout);
    srand(time(0));
    n=rand()%10000;
    a=rand()%4-5;
    b=rand()%20000000-10000000;
    c=rand()%20000000-10000000;
    printf("%d\n%d %d %d\n",n,a,b,c);
    while(n--) printf("%d ",rand()%100+1);
    return 0;
}

checker:

#include<bits/stdc++.h>
using namespace std;
int main()
{
    for(int i=1;;i++)
    {
        printf("The result of No. %d Case is:  ",i);
        system("./datamaker");
        system("./std");
        system("./dp");
        if(system("diff std.out test.out")) {printf("Wrong Answer\n");break;}
        else printf("Accepted\n");
    }
    return 0;
}

最后:

对拍是个技术活,造数据时要考虑数据特性(比如树就需要拿并查集维护),如果时间不是很充裕不要强写对拍,这样只会得不偿失

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