「并查集」程序自动分析

Posted 2022-11-25 nicoppa

程序自动分析

原题链接：程序自动分析

题目大意

给你很多逻辑关系，判断是否有冲突

题目题解

先将两个逻辑分开来看，前一个逻辑存在那么后一个逻辑一定不存在，如果存在那么答案就错误了，可以用并查集维护，详细见代码

//#define fre yes

#include <cstdio>
#include <iostream>
#include <algorithm>

const int N = 1000005;
struct message 
    int x, y, e;
 arr[N];
int ele[N << 1];

namespace Union 
    int par[N << 1];
    inline void init(int n) 
        for (int i = 1; i <= n; i++) 
            par[i] = i;
        
    
    
    inline int find(int x) 
        if(par[x] == x) return par[x];
        else return par[x] = find(par[x]);
    
    
    inline void unite(int x, int y) 
        int a = find(x);
        int b = find(y);
        if(a == b) return ;
        
        par[a] = b;
    
    
    inline bool same(int x, int y) 
        return find(x) == find(y);
    


int n, m;
inline void discrete() 
    std::sort(ele + 1, ele + 1 + n * 2);
    m = std::unique(ele + 1, ele + 1 + 2 * n) - ele - 1;


inline int ask(int x) 
    return std::lower_bound(ele + 1, ele + 1 + m, x) - ele;


int main() 
    static int T;
    scanf("%d", &T);
    while(T--) 
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) 
            scanf("%d %d %d", &arr[i].x, &arr[i].y, &arr[i].e);
            ele[i] = arr[i].x; ele[i + n] = arr[i].y;
         discrete();
        Union::init(m);
        for (int i = 1; i <= n; i++) 
            if(arr[i].e == 1) 
                Union::unite(ask(arr[i].x), ask(arr[i].y));
            
        
        
        bool flag = true;
        for (int i = 1; i <= n; i++) 
            if(arr[i].e == 0) 
                flag = flag && !Union::same(ask(arr[i].x), ask(arr[i].y));
            
        
        
        if(flag) puts("YES");
        else puts("NO");
     return 0;