【Bzoj4195】程序自動分析

題意

給定n個形如xi=xj或xi≠xj的變量相等/不等的約束條件，讓你判斷是否矛盾。

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解析

用並查集把相等的並起來，之後判斷有沒有衝突的。注意要先把所有的都並起來，因爲可能有連等。

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#include <cstdio>
#include <algorithm>
#include <cstring>

#define Rep( i , _begin , _end ) for(int i=(_begin);i<=(_end);i++)
#define For( i , _begin , _end ) for(int i=(_begin);i!=(_end);i++)

using std :: sort;
using std :: max;
using std :: min;
using std :: lower_bound;

const int maxx = 1000000 + 25;

int a[maxx],b[maxx],c[maxx],mp[maxx<<1],ftr[maxx];

int n,m,t,num;
bool flag;

inline int read(){
    int x = 0,f = 1;char c = getchar();
    while(c>'9' || c<'0') {if(f == '-') f = -1; c = getchar();}
    while(c>='0' && c<='9') {x = x * 10 + c-'0';c = getchar();}
    return x * f;
}

void init(){
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    num = n = m = 0;
    flag = false;
}

int find(int x){
    return ftr[x] == x? x : ftr[x] = find(ftr[x]);
}

int main(){
    scanf("%d",&t);

    while( t-- ){
        init();
        scanf("%d",&n);
        Rep( i , 1 , n ){
            a[i] = read();b[i] = read();c[i] = read();
            mp[++num] = a[i];mp[++num] = b[i];
        }
        sort(mp+1,mp+num+1);
        Rep( i , 1 , num ) ftr[i] = i;
        Rep( i , 1 , n ){
            if(c[i] == 1){
                int x = lower_bound(mp+1,mp+num+1,a[i])-mp;
                int y = lower_bound(mp+1,mp+num+1,b[i])-mp;
                int u = find(x),v = find(y);
                if(u != v) ftr[v] = u;
            }
        }

        Rep( i , 1 , n ){
            if(c[i] == 0){
                int x = lower_bound(mp+1,mp+num+1,a[i])-mp;
                int y = lower_bound(mp+1,mp+num+1,b[i])-mp;
                int u = find(x),v = find(y);
                if(u == v) {flag = true;break;}
            }
        }

        puts(flag? "NO" : "YES");
    }
    return 0;
}