Hustoj SPJ example

What is? SPJ

SPJ yes Special Judge It means .

When to use SPJ

When there is more than one answer to the question , We have to use SPJ.

How to use SPJ

Open... In the title SPJ

First , We need to work out the questions , increase SPJ Options , As shown in the figure below .

After the title is saved , It shows that SPJ, As shown in the figure below .

To write SPJ Program

SPJ Program , It's a standard C perhaps C++ Program , According to the requirements of the topic , Read the test file （*.in）, Standard output file （*.out）, User output file （user.out）, Compare .
Use input with parameters .

SPJ Template code

//
#include <bits/stdc++.h>
using namespace std;
using LL=long long;
const int AC=0;
const int WA=1;

int main(int argc,char *args[]) {
    
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	// Read three files , Be careful  arg  Order cannot be changed 
    ifstream f_in;
    f_in.open(args[1]); 
    ifstream f_out;
    f_out.open(args[2]); 
    ifstream f_user;
    f_user.open(args[3]); 
    
    int ret=AC;
/************** Write the logic of the problem **************/
/************************************/
	f_in.close();
	f_out.close();
	f_user.close();
	
	return 0;
}

Upload spj.cc Go to the corresponding topic directory

Suppose our title is $6387$, Then the corresponding path is /home/judge/data/6387. take spj.cc Upload to the corresponding directory , Here's the picture .
spj.cc The file name can be named arbitrarily .

compile spj Executable file

perform

g++ -o spj 6387_spj.cpp -std=c++14

In this way, we can get the corresponding executable file spj.

Modify owner

take spj The owner is changed to www-data.

sudo chown www-data:judge spj

The permissions are set as shown in the following figure .

Set executable

sudo chmod +x spj

test

Build a user.out file .

./spj sample.in sample.out user.out

View the execution results

echo $?

Here's the picture .

Show $0$ The result is correct . Display non-zero is generally $1$ Indicates that the result failed .

SPJ Write sample

SPJ The difficulty of writing lies in the logical judgment of the topic .
We use MYOJ Question no $6387$ As an example .
This problem is a topological sequence of digraphs , Is to give a directed graph , Output the corresponding topology sequence . If it's a $DAG$ Output any reasonable sequence , If not $DAG$ Output $-1$.

Scenario analysis

No $DAG$

We should output $-1$. The user may have two kinds of errors .
Situation 1 ： Output $-12...$. The user output $-1$, But it also outputs multiple data .
Situation two ： Output $-2$. User output non $-1$.

yes $DAG$

We read the user's output , Then, according to the properties of topological sorting , Analyze the legitimacy of each output .
The user may have two kinds of errors .
Situation 1 ： Is not a legitimate topological sort . That is, output to a vertex $u$ When , The summit $u$ The degree of penetration is not $0$.
Situation two ： The user has output a ratio $n$ More vertices .

Sample code

//
#include <bits/stdc++.h>
using namespace std;
using LL=long long;
const int AC=0;
const int WA=1;

const int N=1e5+10;
LL h[N], in[N];
LL ans[N];
const int M=2e5+10;
LL e[M], ne[M], idx;
LL que[M];
LL hh, tt;
LL n;

void add(LL a, LL b) {
    
    e[idx]=b;
    ne[idx]=h[a];
    h[a]=idx++;
    // Update penetration 
    in[b]++;
}

int main(int argc,char *args[]) {
    
#if 1
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
#endif

    ifstream f_in;
    f_in.open(args[1]); 
    ifstream f_out;
    f_out.open(args[2]); 
    ifstream f_user;
    f_user.open(args[3]); 
    
    int ret=AC;
/************** Write the logic of the problem **************/
    memset(h, -1, sizeof h);

    LL m;
    f_in>>n>>m;
    for (LL i=1; i<=m; i++) {
    
        LL a,b;
        f_in>>a>>b;
        add(a,b);
    }
    
    LL cnt=0; 
	LL u;
    while (f_user>>u) {
    
    	//cout<<"u="<<u<<"\n";
    	
    	++cnt;
    	if (cnt>n) {
    
    		return WA+6;
		}

    	if (u>0) {
    
	        if (in[u]!=0) {
    
	        	return WA+1;
			}
	        // Traverse u All sides of 
	        for (LL i=h[u]; i!=-1; i=ne[i]) {
    
	            LL v=e[i];
	            in[v]--;
	        }    		
		} else if (u!=-1) {
    
			return AC+2;
		}
    }
    if (cnt!=n) {
    
    	return WA+5;
	}

/************************************/
	f_in.close();
	f_out.close();
	f_user.close();
	
	return 0;
}