Description
There are three types of animals in the animal kingdom, A, B, and C. The food chain of these three types of animals forms an interesting ring. A eats B, B eats C, and C eats A.
There are currently N animals, numbered from 1 to N. Each animal is one of A, B, and C, but we don't know which one it is.
Some people describe the food chain relationship formed by these N animals in two ways: the
first one is "1 X Y", which means that X and Y are of the same kind.
The second argument is "2 X Y", which means X eats Y.
This person uses the above two statements to say K sentences to N animals one after another. Some of these K sentences are true and some are false. When a sentence satisfies one of the following three items, the sentence is false, otherwise it is true.
1) The current words conflict with some previous true words, which is a lie;
2) The current words X or Y is greater than N, which is a lie;
3) The current words mean that X eats X, which is a lie.
Your task is to output the total number of lies based on a given N (1 <= N <= 50,000) and K sentences (0 <= K <= 100,000).
Input
The first line is two integers N and K, separated by a space.
Each of the following K lines contains three positive integers D, X, Y, separated by a space between the two numbers, where D represents the type of argument.
If D=1, it means that X and Y are of the same kind.
If D=2, it means X eats Y.
Output
There is only one integer, which represents the number of lies.
Sample Input
100 7
1 101 1
2 1 2
2 2 3
2 3 3
1 1 3
2 3 1
1 5 5
Sample Output
3
Ideas
Use a 3*N size array to correspond to the three animals A, B, and C respectively. Use and search for different relationships to establish contacts. If conflicts occur, record them. (See code comments for details)
Code
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<string>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
//#include<bits/stdc++.h>
#define pb push_back
using namespace std;
const int maxn=1e6+5;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int minn=0xc0c0c0c0;
int n,k,x[maxn],y[maxn],r[maxn],pre[maxn];
int find(int x)
{
if(x==pre[x])
return x;
else
return pre[x]=find(pre[x]);
}
void unite(int a,int b)
{
int fa=find(a);
int fb=find(b);
if(fa!=fb)
pre[fa]=fb;
}
bool judge(int a,int b)
{
return find(a)==find(b);
}
int main()
{
scanf("%d%d",&n,&k);
for(int i=0;i<k;i++)
scanf("%d%d%d",&r[i],&x[i],&y[i]);//说法种类，编号
for(int i=0;i<=n;i++)
pre[i]=i;
int ans=0;
for(int i=0;i<k;i++)
{
int num=r[i];
int a=x[i]-1,b=y[i]-1;
if(a<0||a>n-1||b<0||b>n-1)
{
ans++;
continue;
}//条件越界
if(num==1)//a和b是同类
{
if(judge(a,b+n)||judge(a,b+2*n))
ans++;//非同类动物在同一集合中，产生矛盾
else
{
unite(a,b);//A类合并
unite(a+n,b+n);//B类合并
unite(a+2*n,b+2*n);//C类合并
}
}
else
{
if(judge(a,b)||judge(a,b+2*n))
ans++;//a,b已是同类或出现逆食物链的情况
else
{
unite(a,b+n);//A吃B
unite(a+n,b+2*n);//B吃C
unite(a+2*n,b);//C吃A
}
}
}
printf("%d\n",ans);
return 0;
}