Codeforces 101572 D - Distinctive Character
思路:bfs
使最大的匹配数最小,转换一下,就是使最小的不匹配数最大,用bfs找最大的距离
代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pii pair<int, int> #define piii pair<pii, int> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout); //head const int N = 1e5 + 5; queue<int> q; int dis[N*15]; int main() { fio; int n, k, ans; string s; cin >> n >> k; mem(dis, -1); for (int i = 1; i <= n; i++) { cin >> s; int now = 0; for (int j = 0; j < k; j++) if(s[j] == '1') now |= 1<<j; q.push(now); dis[now] = 0; ans = now; } int mx = 0; while(!q.empty()) { int now = q.front(); q.pop(); for (int i = 0; i < k; i++) { int nxt = now^(1<<i); if(dis[nxt] == -1) { dis[nxt] = dis[now] + 1; q.push(nxt); if(dis[nxt] > mx) { mx = dis[nxt]; ans = nxt; } } } } for (int i = 0; i < k; i++) { if(ans & (1<<i)) cout << 1; else cout << 0; } cout << endl; return 0; }