Codeforces 468b two sets and query Sets

The N number is given. The N number must be allocated to two sets. The element x in the Set 0 must be a-X and then 0, which is the same as the 1 set.

I have written several versions, and Wa has been in the 8th sets of data. At last, I referred to ANS and wrote and checked the set.

Learn: 1. Pay attention to discrete logic thinking. The official answer is to push the inverse negative proposition from the condition.

2. query the set: Fa [find (I)] = MP [A-P [I]? Find (a-p [I]): Find (n + 2 );

3. The method of discretization and then hash can be easily written with map time.

//#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <iostream>#include <iomanip>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;#define ls(rt) rt*2#define rs(rt) rt*2+1#define ll long long#define ull unsigned long long#define rep(i,s,e) for(int i=s;i<e;i++)#define repe(i,s,e) for(int i=s;i<=e;i++)#define CL(a,b) memset(a,b,sizeof(a))#define IN(s) freopen(s,"r",stdin)#define OUT(s) freopen(s,"w",stdout)const ll ll_INF = ((ull)(-1))>>1;const double EPS = 1e-8;const double pi = acos(-1.0);const int INF = 100000000;const int MAXN = 1e5+100;int fa[MAXN],p[MAXN],vis[MAXN];int n,a,b;map <int , int> mp;int Fi(int x){    if(x!=fa[x])fa[x]=Fi(fa[x]);    return fa[x];}void un(int x, int y){    x=Fi(x);    y=Fi(y);    fa[x]=y;}int flag;void init(){    for(int i=0;i<=n+3;i++)        fa[i]=i;    mp.clear();    CL(vis,0xff);    flag=1;}/*void solve(){    for(int i=1;i<=n;i++)    {        if(mp[a-p[i]])        {            if(Fi(fa[mp[a-p[i]]])==Fi(n+2) || mp[b-p[i]])            {                flag=0;                return;            }            else            {                un(mp[a-p[i]],n+1);            }        }        else        {            if(mp[b-p[i]])            {                if(Fi(fa[mp[b-p[i]]]) == Fi(n+1) || mp[a-p[i]])                {                    flag=0;                   return;                }                else                {                    un(mp[b-p[i]],n+2);                }            }            else            {                flag=0;                return;            }        }    }}*/void solve(){    for(int i=1;i<=n;i++)    {        fa[Fi(i)] = mp[a-p[i]] ? Fi(mp[a-p[i]]) : Fi(n+2);        fa[Fi(i)] = mp[b-p[i]] ? Fi(mp[b-p[i]]) : Fi(n+1);    }}int main(){    //IN("D.txt");    while(~scanf("%d%d%d",&n,&a,&b))    {        init();        for(int i=1;i<=n;i++)        {            scanf("%d",&p[i]);            mp[p[i]]=i;        }        solve();        if(flag && Fi(n+1)!=Fi(n+2))        {            printf("YES\n");            int ans[MAXN],cnt=1;            for(int i=1;i<=n;i++)            {                if(Fi(i) == Fi(n+1))                    ans[i]=0;                else                    ans[i]=1;            }            printf("%d",ans[1]);            for(int i=2;i<=n;i++)                printf(" %d",ans[i]);            putchar('\n');        }        else            puts("NO");    }    return 0;}

Codeforces 468b two sets and query Sets