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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#260578 | #7738. Equivalent Rewriting | ucup-team1447# | WA | 11ms | 97252kb | C++14 | 3.0kb | 2023-11-22 13:17:06 | 2023-11-22 13:17:08 |
Judging History
answer
// what is matter? never mind.
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,sse4,popcnt,abm,mmx,avx,avx2")
#include<bits/stdc++.h>
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
#define ll long long
#define ull unsigned long long
#define int long long
using namespace std;
inline int read()
{
char c=getchar();int x=0;bool f=0;
for(;!isdigit(c);c=getchar())f^=!(c^45);
for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
if(f)x=-x;return x;
}
#define mod 1000000007
struct modint{
int x;
modint(int o=0){x=o;}
modint &operator = (int o){return x=o,*this;}
modint &operator +=(modint o){return x=x+o.x>=mod?x+o.x-mod:x+o.x,*this;}
modint &operator -=(modint o){return x=x-o.x<0?x-o.x+mod:x-o.x,*this;}
modint &operator *=(modint o){return x=1ll*x*o.x%mod,*this;}
modint &operator ^=(int b){
modint a=*this,c=1;
for(;b;b>>=1,a*=a)if(b&1)c*=a;
return x=c.x,*this;
}
modint &operator /=(modint o){return *this *=o^=mod-2;}
friend modint operator +(modint a,modint b){return a+=b;}
friend modint operator -(modint a,modint b){return a-=b;}
friend modint operator *(modint a,modint b){return a*=b;}
friend modint operator /(modint a,modint b){return a/=b;}
friend modint operator ^(modint a,int b){return a^=b;}
friend bool operator ==(modint a,int b){return a.x==b;}
friend bool operator !=(modint a,int b){return a.x!=b;}
bool operator ! () {return !x;}
modint operator - () {return x?mod-x:0;}
bool operator <(const modint&b)const{return x<b.x;}
};
inline modint qpow(modint x,int y){return x^y;}
vector<modint> fac,ifac,iv;
inline void initC(int n)
{
if(iv.empty())fac=ifac=iv=vector<modint>(2,1);
int m=iv.size(); ++n;
if(m>=n)return;
iv.resize(n),fac.resize(n),ifac.resize(n);
For(i,m,n-1){
iv[i]=iv[mod%i]*(mod-mod/i);
fac[i]=fac[i-1]*i,ifac[i]=ifac[i-1]*iv[i];
}
}
inline modint C(int n,int m){
if(m<0||n<m)return 0;
return initC(n),fac[n]*ifac[m]*ifac[n-m];
}
inline modint sign(int n){return (n&1)?(mod-1):(1);}
#define fi first
#define se second
#define pb push_back
#define mkp make_pair
typedef pair<int,int>pii;
typedef vector<int>vi;
#define maxn 1000005
#define inf 0x3f3f3f3f
int n,m,res[maxn];
vi a[maxn];
set<int>ban[maxn];
vi p[maxn];
void work()
{
m=read(),n=read();
For(i,1,n)a[i].clear();
For(i,1,m){
ban[i].clear();
res[i]=i;
p[i].resize(read());
for(int&x:p[i]){
x=read();
a[x].pb(i);
}
}
For(i,1,n){
if(a[i].size()){
int sz=a[i].size();
int u=a[i].back();
For(j,0,sz-2){
ban[u].insert(a[i][j]);
ban[a[i][j]].insert(u);
}
}
}
For(i,1,m)
For(j,1,m)
if(i!=j && !ban[i].count(j)) {
swap(res[i],res[j]);
puts("Yes");
For(k,1,m)cout<<res[k]<<" \n"[k==m];
return;
}
puts("No");
}
signed main()
{
int T=read();
while(T--)work();
return 0;
}
/*
*/
详细
Test #1:
score: 0
Wrong Answer
time: 11ms
memory: 97252kb
input:
3 3 6 3 3 1 5 2 5 3 2 2 6 2 3 3 1 3 2 2 3 1 1 3 2 2 1
output:
Yes 3 2 1 No No
result:
wrong answer two transactions are not equivalent. (test case 1)