Codeforces 468b two sets (Bipartite Graph Matching)

Link: codeforces 468b Two 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.

Solution: similar to bipartite graph matching.

#include <cstdio>#include <cstring>#include <map>#include <stack>#include <algorithm>using namespace std;const int maxn = 1e5 + 5;int N, A, B, x[maxn], v[maxn], r = 0;map<int, int> G;bool match (int a, int M, int k) {    int p = G[a];    if (!G.count(M - a))        return false;    int q = G[M - a];    if (v[q] == -1 || a * 2 == M) {        v[p] = v[q] = k;    } else {        if (match(A + B - 2 * M + a, M, k))            v[p] = v[q] = k;        else            return false;    }    return true;}bool solve () {    if (r >= max(A,B))        return false;    for (int i = 1; i <= N; i++) {        if (v[i] != -1)            continue;        if (!match(x[i], A, 0) && !match(x[i], B, 1))            return false;    }    return true;}int main () {    scanf("%d%d%d", &N, &A, &B);    memset(v, -1, sizeof(v));    for (int i = 1; i <= N; i++) {        scanf("%d", &x[i]);        r = max(x[i], r);        G[x[i]] = i;    }    if (solve()) {        printf("YES\n");        for (int i = 1; i <= N; i++)            printf("%d%c", v[i], i == N ? ‘\n‘ : ‘ ‘);    } else        printf("NO\n");    return 0;}

Codeforces 468b two sets (Bipartite Graph Matching)