Codeforces 101572 D - Distinctive Character

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 + ;
queue<int> q;
int dis[N*];
int main() {
fio;
int n, k, ans;
string s;
cin >> n >> k;
mem(dis, -);
for (int i = ; i <= n; i++) {
cin >> s;
int now = ;
for (int j = ; j < k; j++) if(s[j] == '') now |= <<j;
q.push(now);
dis[now] = ;
ans = now;
}
int mx = ;
while(!q.empty()) {
int now = q.front();
q.pop();
for (int i = ; i < k; i++) {
int nxt = now^(<<i);
if(dis[nxt] == -) {
dis[nxt] = dis[now] + ;
q.push(nxt);
if(dis[nxt] > mx) {
mx = dis[nxt];
ans = nxt;
}
}
}
}
for (int i = ; i < k; i++) {
if(ans & (<<i)) cout << ;
else cout << ;
}
cout << endl;
return ;
}
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