题意:
n distinct integers: p1, p2, ..., pn. He wants to divide all of them into two sets A and B. The following two conditions must be satisfied:
- xbelongs to setA, then numbera-xmust also belong to setA.
- xbelongs to setB, then numberb-xmust also belong to setB.
Help Little X divide the numbers into two sets or determine that it's impossible.
Input
n, a, b (1 ≤ n ≤ 105; 1 ≤ a, b ≤ 109). The next line contains n space-separated distinct integers p1, p2, ..., pn (1 ≤ pi ≤ 109).
Output
YES" in the first line. Then print n integers: b1, b2, ..., bn (bi equals either 0, or 1), describing the division. If bi equals to 0, then pi belongs to set A, otherwise it belongs to set B.
NO" (without the quotes).
题解:
因为一个数要么A要么属于B..那么将它们都先定为A..再把不符合的调整到B..由于改变了一个位置从A到B..会让若干的数也会变..用一个dfs来处理就好...
Program:
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<string>
#include<queue>
#include<map>
#include<algorithm>
#define MAXN 100005
#define ll long long
#define oo 1e+10
#define eps 1e-10
using namespace std;
struct node
{
int w,id;
bool operator<(node a) const
{
return w<a.w;
}
}P[MAXN];
int ans[MAXN],n,a,b;
int find(int x)
{
int l,r,mid;
l=1,r=n+1;
while (r-l>1)
{
mid=r+l>>1;
if (P[mid].w>x) r=mid;
else l=mid;
}
if (P[l].w!=x) return 0;
return l;
}
bool dfs(int x)
{
if (!x) return false;
if (ans[P[x].id]) return true;
ans[P[x].id]=1;
if (find(a-P[x].w))
if (!dfs(find(a-P[x].w))) return false;
if (!dfs(find(b-P[x].w))) return false;
return true;
}
bool FindAns()
{
int i;
memset(ans,0,sizeof(ans));
for (i=1;i<=n;i++)
{
if (find(a-P[i].w)) continue;
if (!dfs(i)) return false;
}
puts("YES");
for (i=1;i<n;i++) printf("%d ",ans[i]);
printf("%d\n",ans[n]);
return true;
}
int main()
{
int i;
scanf("%d%d%d",&n,&a,&b);
for (i=1;i<=n;i++) scanf("%d",&P[i].w),P[i].id=i;
sort(P+1,P+n+1);
if (!FindAns()) puts("NO");
return 0;
}
