OJ Board

dfs()
for each person ,
assume he is speaking truth,
-> All reliables are speaking truth
dfs(for each reliable)
if he is false
do nothing.

As an example, let's assume there are four informants A, B, C and D, with the following surveyed answers: ``A says B is reliable but D is not'', ``B says C is not reliable'', and ``C says A and D are reliable''. In this case, it happens that at most two informants are reliable.

tryit1 wrote:

As an example, let's assume there are four informants A, B, C and D, with the following surveyed answers: ``A says B is reliable but D is not'', ``B says C is not reliable'', and ``C says A and D are reliable''. In this case, it happens that at most two informants are reliable.

Can you tell me how we get 2 informans as reliable ? how would you approach the above example. Any similar problems ?

Ithink it is a easy problem

But I got WA........

Some one PLZ check my code.........

#include<stdio.h>
#include<stdlib.h>

int a[803];

int main()
{
	int i,n,b,c,m,Count,j,d;
	while(scanf("%d%d",&m,&n)==2 && (m || n))
	{
		Count= 0;
		for(i=1;i<=m;i++)
			a[i]=1;
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&b,&c);

			if(a[b])
			{
				if( c > 0)
				{
					a[c] = 1;
					
				}
				else if(c < 0)
				{
					d = (-1)*c;
					a[d] = 0;
				}			
			}
		}
		for(i=1;i<=m;i++)
		{
			if(a[i])
				Count++;
		}
		printf("%d\n",Count);
	}
	return 0;
}

Thanks hasan3050 for ur test case
I solve this test case But again WA
My modified code Here:
PLZ help me.....

#include<stdio.h>
#include<stdlib.h>

int a[803];

int main()
{
	int i,n,b,c,m,Count,d,e;
	//freopen("a.txt","r",stdin);
	while(scanf("%d%d",&m,&n)==2 && (m || n))
	{
		Count= 0;
		for(i=1;i<=m;i++)
			a[i]=1;
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&b,&c);
			e = c;
			if( e < 0)
			{
				e = (-1)*e;
			}
			if( (a[b]==2) && ( a[e] > 0) )
			{
				if( c > 0)
				{
					a[c] = 2;
					
				}
				else if(c < 0)
				{					
					a[e] = 0;
				}			
			}
			else if( (a[b]==1) && ( a[e] > 0) )
			{
				if( c > 0)
				{
					a[c] = 2;
					
				}
				else if(c < 0)
				{					
					a[e] = 0;
				}
			}
		}

		for(i=1;i<=m;i++)
		{
			//printf("a [%d] == %d\n",i,a[i]);
			if(a[i]>=1)
				Count++;
		}
		printf("\n%d\n",Count);
	}
	return 0;
}

tryit1 wrote:how to solve this problem ?

Can you post links to similar problems.

Igor9669 wrote:First you print leading new line!
Second why do you use such a big array you need only 20 elements and third try this test:

Code: Select all

3 2
1 -3
3 -2
0 0

Answer should be 1!

Jehad Uddin wrote:because 3 isnt reliable and 2 isnt reliable ,so only 1 is reliable so ans is 1,its so simple problem,try all the I/Os in the board,