51 nod 1557 两个集合

只需要将满足条件的数尽可能的放一个集合里面，然后看有没有不满足的条件发生，注意可能先放A也可能先放B，所以要跑两遍，一边先放A，一边先放B(没注意到这点wa了好多发。)

#include<bits/stdc++.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

typedef long long ll;
typedef double dl;

const int N = 1e5+7;
const int M = 1e9+7;
const int INF = 0x7fffffff;

int p[N];
set<int> st;
set<int> sta;
set<int> stb;
set<int> st1;
set<int> sta1;
set<int> stb1;
void solve()
{
    int n,A,B;
    scanf("%d%d%d",&n,&A,&B);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&p[i]);
        st.insert(p[i]);
        st1.insert(p[i]);
    }
    sort(p+1,p+n+1);
    int flag1=1;
    for(int i=n;i>=1;i--)
    {
        if(st.count(p[i])==0) continue;
        else if(st.count(A-p[i])||sta.count(p[i])||sta.count(A-p[i]))
        {
            if(st.count(p[i]))
            sta.insert(p[i]);
            if(st.count(A-p[i]))
            sta.insert(A-p[i]);
            if(st.count(p[i]))
            st.erase(p[i]);
            if(st.count(A-p[i]))
            st.erase(A-p[i]);
        }
        else if(st.count(B-p[i])||stb.count(p[i])||stb.count(B-p[i]))
        {
            if(st.count(p[i]))
            stb.insert(p[i]);
            if(st.count(B-p[i]))
            stb.insert(B-p[i]);
            if(st.count(p[i]))
            st.erase(p[i]);
            if(st.count(B-p[i]))
            st.erase(B-p[i]);
        }
        else flag1=0;
    }
    int flag2=1;
    for(int i=n;i>=1;i--)
    {
        if(st1.count(p[i])==0) continue;
        else if(st1.count(B-p[i])||sta1.count(p[i])||sta1.count(B-p[i]))
        {
            if(st1.count(p[i]))
            sta1.insert(p[i]);
            if(st1.count(B-p[i]))
            sta1.insert(B-p[i]);
            if(st1.count(p[i]))
            st1.erase(p[i]);
            if(st1.count(B-p[i]))
            st1.erase(B-p[i]);
        }
        else if(st1.count(A-p[i])||stb1.count(p[i])||stb1.count(A-p[i]))
        {
            if(st1.count(p[i]))
            stb1.insert(p[i]);
            if(st1.count(A-p[i]))
            stb1.insert(A-p[i]);
            if(st1.count(p[i]))
            st1.erase(p[i]);
            if(st1.count(A-p[i]))
            st1.erase(A-p[i]);
        }
        else flag2=0;
    }
    if(flag1||flag2)
    {
        printf("YES\n");
    }
    else printf("NO\n");

}

int main()
{
    solve();
}