Codeforces 468B sets and set deformation

Title Link: Http://codeforces.com/problemset/problem/468/B

The main topic: give you n different integers, and then give two integers, a, B, ask whether the n number can be divided into two parts, so that in a this set of any number of AI, there are A-ai also in a this set, the same with the B set. If there is no reasonable division method, output no, otherwise output yes, and lose to a reasonable grouping.

Problem Solving Ideas:

First we assume that there is a reasonable scheme, then there is one for either number x, which is definitely in the A/b collection. At the same time, if the scheme is reasonable, the final result is no intersection of the elements in the set.

1. For a number x, if the a-x does not exist, then x must be in Group B (assuming the group is formed), then the X and B are grouped together, if at the same time the b-x also does not exist, then the X and a are grouped together, at this time A and B are grouped together, obviously this scheme is unreasonable.

2. For a number x, if the a-x exists, then x and a-x corresponding subscript together, it is not possible to say that x belongs to set a, because there may be n=2,a=8,b=20,4,16, such data, only if the b-x does not exist, only the X and a subscript together.

3, if the b-x also exist, the X and B-x also the subscript also up, at this time if X. =a-x. =b-x, it is also unreasonable to note that A and B sets are common x, that is, the scheme A and B are combined. At this point if x==a-x. =b-x, obviously should be grouped into B, of course, the exception is a and b are equal, at this time any grouping can be.

The overall situation should be these kinds.

For the code, it feels good to have the following points:

1,map mapping makes it easier to find pairs of numbers.

2, virtual out two A, a set of root nodes n+1 and n+2;

3, and check the special treatment of the set;

The code is as follows:

#include <stdio.h> #include <cstring> #include <map> #include <algorithm> using namespace std;
#define N 100005 int f[n],num[n];
Map <int,int> MP;
    int find (int x) {return x!=f[x]?f[x]=find (F[x]): f[x];} void merge (int x,int y) {int fx=find (x);
    int Fy=find (y);
    if (fx!=fy) f[fx]=fy;
Return
    } int main () {int n,a,b;
    scanf ("%d%d%d", &n,&a,&b);
    for (int i=1;i<=n+2;i++) f[i]=i;
        for (int i=1;i<=n;i++) {scanf ("%d", &num[i]);
    Mp[num[i]]=i;
        } for (int i=1;i<=n;i++) {if (Mp[a-num[i]) merge (I,mp[a-num[i]);
        else merge (i,n+2);
        if (Mp[b-num[i]]) merge (I,mp[b-num[i]]);
    else merge (i,n+1);
    } if (Find (n+1) ==find (n+2)) printf ("no\n");
        else {printf ("yes\n");
            for (int i=1;i<=n;i++) {if (Find (i) ==find (n+1)) printf ("0");
        else printf ("1");
    } printf ("\ n"); }
    return 0;
 }