题目描述

思路

  • 首先这个题目判断顺序不会影响最终答案,可以先判断所有等式,再判断所有不等式,只要有不等式不成立那就是错误。(相等:在同一集合,不相等:在不同集合)
  • 因为数据范围是1E9,会超过限制内存。我们需要做一个哈希来映射。

代码

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import java.io.*;
import java.util.*;

class Main{
public static int N = 100010,P = N*2 - 1;
public static Option[] os = new Option[N];
public static int[] hash = new int[N*2],find = new int[N*2];
public static Scanner sc = new Scanner(System.in);

public static int get(int x){
int k = x % P;
for(int i = k;i < N * 2 - 1;i ++){
if(hash[i] == -1 || hash[i] == x)
return i;
}
return -1;
}

public static int find(int x){
if(find[x] != x) find[x] = find(find[x]);
return find[x];
}

public static void main(String[] args){
int k = sc.nextInt();
while((k--)!=0){
for(int i = 1;i < N*2;i++) find[i] = i;
Arrays.fill(hash,-1);
int n = sc.nextInt();
for(int i = 1;i <= n;i++){
int x1 = sc.nextInt(),x2 = sc.nextInt(),o = sc.nextInt();
os[i] = new Option(x1,x2,o);
}

//等号条件都加进去

for(int i = 1;i <= n;i++){
if(os[i].o == 1)
{
int x1 = os[i].x1,x2 = os[i].x2;
int a = get(x1),b = get(x2);
hash[a] = x1;
hash[b] = x2;
int pa = find(a),pb = find(b);
if(pa != pb) find[pa] = pb;

}
}

int index = 0;
//判断不等号
for(int i = 1;i <= n;i++){
if(os[i].o == 0){
int x1 = os[i].x1,x2 = os[i].x2;
int a = get(x1),b = get(x2);
int pa = find(a),pb = find(b);
if(pa == pb){
index = 1;
System.out.println("NO");
break;
}
}
}

if(index == 0) System.out.println("YES");
}
}


static class Option{
int x1,x2,o;
public Option(int x1,int x2,int o){
this.x1 = x1;
this.x2 = x2;
this.o = o;
}

}
}