Codeforces Round #268 (Div. 2) D. Two Sets
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【题意】给了N个数,和a,b,现在要把这N个数放进两个集合,如果x属于集合A,必须满足x-a也属于A集合,B同理。问能否满足这个条件,不能输出No,能就输出Yes,并且输出这个数属于哪个集合,B就输出1,A就输出了。
【解题方法】可以用并查集的思想来处理,设置额外的两个元素n+1和n+2。使得n+1是A元素,n+2属于B集合。
【AC 代码】
#include <map>#include <cstdio>#include <cstring>#include <iostream>using namespace std;int fa[100010],s[100010];map<int,int>mp;int n,a,b;int Find(int x){ if(x==fa[x]) return x; else return fa[x]=Find(fa[x]);}int main(){ cin>>n>>a>>b; for(int i=1; i<=n; i++){ cin>>s[i]; mp[s[i]]=i; } for(int i=1; i<=n+2; i++) fa[i]=i; for(int i=1; i<=n; i++){ int t=Find(i); if(mp[a-s[i]]) fa[t]=Find(mp[a-s[i]]); else fa[t]=Find(n+2);//代表B集合 int r=Find(i); if(mp[b-s[i]]) fa[r]=Find(mp[b-s[i]]); else fa[r]=Find(n+1);//代表A集合 } if(Find(n+1)==Find(n+2)){puts("NO");} else{ puts("YES"); for(int i=1; i<n; i++){ printf("%d ",Find(i)==Find(n+2)); } printf("%d\n",Find(n)==Find(n+2)); }}
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