P1955 程序自动分析

Posted 2022-03-14 ukcxrtjr

题面:https://www.luogu.org/problemnew/show/P1955

本题可以把等于的并在一起,碰到不等于的就判断是否矛盾。由于数据比较大，注意要离散化。

Code:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<ctime>
using namespace std;
const int N=1000005;
int t,n,len,b[N],fa[N];
bool flag;
struct Node
    int i,j,e;
a[N];
bool cmp(Node a,Node b)
    return a.e>b.e;

int find(int x)
    if(fa[x]==x)
        return fa[x];
    
    return fa[x]=find(fa[x]);

void merge(int a,int b)
    int x=find(a),y=find(b);
    if(x!=y)
        fa[x]=y;
    

int main()
    scanf("%d",&t);
    while(t--)
        scanf("%d",&n);
        len=0;
        for(int i=1;i<=n;i++)
            scanf("%d%d%d",&a[i].i,&a[i].j,&a[i].e);
            b[++len]=a[i].i;
            b[++len]=a[i].j;
        
        sort(a+1,a+1+n,cmp);
        sort(b+1,b+1+len);
        len=unique(b+1,b+len+1)-(b+1);
        for(int i=1;i<=len;i++)
            fa[i]=i;
        
        flag=true;
        for(int i=1;i<=n;i++)
            a[i].i=lower_bound(b+1,b+len+1,a[i].i)-b;
            a[i].j=lower_bound(b+1,b+len+1,a[i].j)-b;
            if(a[i].e)
                merge(a[i].i,a[i].j);
            
            else if(find(a[i].i)==find(a[i].j))
                flag=false;
                break;
            
        
        if(flag)
            printf("YES\n");
        
        else
            printf("NO\n");
        
    
    return 0;