240. Food Chain
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 statement is 1 XY, which means that X and Y are of the same kind.
The second argument is 2 XY, which means X eats Y.
This person uses the above two statements to say K sentences to N animals. 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.
The current words conflict with some of the previous true words, which is a lie; in the
current words X or Y is greater than N, it is a lie; the
current words mean that X eats X, which is a lie.
Your task is to output the total number of false words based on the given N and K sentences.
Input format 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.
The output format
has only one integer, which represents the number of lies.
Data range
1≤N≤50000,
0≤K≤100000
Input sample:
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
Code
#include <iostream>
#define read(x) scanf("%d",&x)
using namespace std;
const int N=5e4+10;
int pre[N],dis[N];
int find(int x)
{
if (pre[x]!=x) {
int t=pre[x];
pre[x]=find(pre[x]);
dis[x]= (dis[x]+dis[t])%3;
}
return pre[x];
}
int main()
{
int n,k;
read(n),read(k);
//初始化并查集，动物编号从1开始
for (int i=1;i<=N;i++) pre[i]=i;
//开始输入并判断
int s,x,y;
int res=0;
while (k--) {
read(s),read(x),read(y);
s--;//我们规定的状态距离为0表示同类，距离为1表示x吃y，距离为2表示y吃x
//输入的s为1表示同类，2表示x吃y，让s代替距离
if ( x>n||y>n||(s&&(x==y)) ) {
res++; continue;}
int px=find(x),py=find(y);
if (px!=py) {
//x所在加入集合中y所在集合中
pre[px]=py; dis[px]=(dis[y]+s-dis[x])%3; //根据输入，要么是x吃y，要么是x和y同类，认为dis[x]>=dis[y]
} else {
//等式dis[x]+dis[px]-s和dis[y]模3同余
if ((dis[x]-dis[y]-s)%3) res++;
}
}
printf("%d",res);
return 0;
}