BZOJ 4195: [Noi2015]程序自动分析 [并查集 离散化 | 种类并查集WA]

题意：

给出若干相等和不等关系，判断是否可行

---

woc NOI考这么傻逼的题飞快打了一个种类并查集交上了然后爆零...

发现相等和不等看错了异或一下再叫woc90分

然后发现md$a \neq b, a \neq c,不能得到b = c$

老老实实的把所有相等关系加并查集然后不等关系来判断吧，唉

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=2e6+;
typedef long long ll;
inline int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x*f;
}

int n, mp[N], m;
struct meow{
    int x, y, e;
    bool operator <(const meow &a) const {return e<a.e;}
} a[N];
int fa[N], val[N];
int find(int x) {return x==fa[x] ? x : fa[x]=find(fa[x]);}

int main() {
    freopen("in","r",stdin);
    int T=read();
    while(T--) {
        n=read(); m=;
        for(int i=; i<=n; i++)
            mp[++m]=a[i].x=read(), mp[++m]=a[i].y=read(), a[i].e=read()^;
        sort(mp+, mp++m); m = unique(mp+, mp++m) - mp - ;
        for(int i=; i<=m; i++) fa[i]=i, val[i]=;
        int flag=, i;
        sort(a+, a++n);
        for(i=; i<=n && !a[i].e; i++) {
            a[i].x = lower_bound(mp+, mp++m, a[i].x) - mp;
            a[i].y = lower_bound(mp+, mp++m, a[i].y) - mp;
            fa[find(a[i].x)] = find(a[i].y);
        }
        for(; i<=n; i++) {
            a[i].x = lower_bound(mp+, mp++m, a[i].x) - mp;
            a[i].y = lower_bound(mp+, mp++m, a[i].y) - mp;
            if(find(a[i].x) == find(a[i].y) ) {puts("NO"); flag=; break;}
        }
        if(flag) puts("YES");
    }
}