Swapping Characters CodeForces

大意: 给定k个字符串, 长度均为n, 求是否存在一个串S, 使得k个字符串都可以由S恰好交换两个字符得到.

暴力枚举交换的两个字符的位置, 计算出交换后与其他串不同字符的个数, 若为1或>2显然不成立, 若为0必须要求存在两个相同的字符.

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

string s[N];
int n, k, vis[N], d[N];
map<int,int> q[N];

int main() {
	cin>>k>>n;
	REP(i,1,k) {
		cin>>s[i];
		for (auto t:s[i]) if (++q[i][t]>1) vis[i]=1;
		if (q[i]!=q[1]) return puts("-1"),0;
	}
	if (k==1) {
		swap(s[1][0],s[1][1]);
		cout<<s[1]<<endl;
		return 0;
	}
	REP(i,2,k) {
		REP(j,0,n-1) d[i]+=s[1][j]!=s[i][j];
	}
	REP(i,0,n-1) REP(j,i+1,n-1) {
		int ok = 1;
		REP(kk,2,k) {
			int t = d[kk];
			d[kk] -= s[1][i]!=s[kk][i];
			d[kk] -= s[1][j]!=s[kk][j];
			d[kk] += s[1][j]!=s[kk][i];
			d[kk] += s[1][i]!=s[kk][j];
			if (d[kk]==1) ok = 0;
			else if (d[kk]==0&&!vis[kk]) ok = 0;
			else if (d[kk]>2) ok = 0;
			d[kk] = t;
		}
		if (ok) {
			swap(s[1][i],s[1][j]);
			return cout<<s[1]<<endl,0;
		}
	}
	puts("-1");
}