【Bzoj4195】程序自动分析

来源：互联网发布：联通系统集成.知乎编辑：程序博客网时间：2024/05/02 05:01

题意

给定n个形如xi=xj或xi≠xj的变量相等/不等的约束条件，让你判断是否矛盾。

解析

用并查集把相等的并起来，之后判断有没有冲突的。注意要先把所有的都并起来，因为可能有连等。

#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;}

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