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2021百度之星程序设计大赛-初赛一部分题目总结
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部分題目 1001迷失 1003鴿子 1004萌新 1006毒瘤數(shù)據(jù)結(jié)構(gòu)題 1008獵人殺
1001迷失
題目意思是要從11 1 kk k nn n nn n 鄰接矩陣k次方的結(jié)論
如果從ii i jj j a[i][j]=1a[i][j]=1 a [ i ] [ j ] = 1 ak[i][j]a^k[i][j] a k [ i ] [ j ] ii i jj j kk k 拆點(diǎn)法
將每一個(gè)點(diǎn)拆成兩個(gè)點(diǎn)，分別為有附魔的和沒(méi)附魔的，可以分別表示為ii i i+ni+n i + n 上面表示帶附魔的，如果ii i 這樣以后只需要填充鄰接矩陣先計(jì)算出所有點(diǎn)的度數(shù)，接下來(lái)根據(jù)每個(gè)點(diǎn)的的度數(shù)計(jì)算對(duì)模的逆元，之后進(jìn)行矩陣快速冪計(jì)算即可 # include <iostream> # include <cstring> # include <algorithm> # include <queue> # include <stack> # include <vector> # include <cmath> # include <cstdio> # include <map> using namespace std
;
typedef long long ll
;
const int MAXN
= 333 ;
const ll MOD
= 998244353 ;
inline int read ( ) { int x
= 0 , f
= 1 ; char c
= getchar ( ) ; while ( c
< '0' || c
> '9' ) { if ( c
== '-' ) f
= - 1 ; c
= getchar ( ) ; } while ( c
>= '0' && c
<= '9' ) { x
= ( x
<< 1 ) + ( x
<< 3 ) + ( c
^ 48 ) ; c
= getchar ( ) ; } return x
* f
;
}
struct Matrix { ll m
[ MAXN
] [ MAXN
] ; Matrix ( ) { memset ( m
, 0 , sizeof m
) ; }
} ;
Matrix
mul ( Matrix a
, Matrix b
, int n
) { Matrix ans
; for ( int i
= 1 ; i
<= n
; i
++ ) { for ( int j
= 1 ; j
<= n
; j
++ ) { for ( int k
= 1 ; k
<= n
; k
++ ) { ans
. m
[ i
] [ j
] = ( ans
. m
[ i
] [ j
] + a
. m
[ i
] [ k
] * b
. m
[ k
] [ j
] ) % MOD
; } } } return ans
;
}
Matrix
fastpow ( Matrix base
, int power
, int n
) { Matrix ans
; for ( int i
= 1 ; i
<= n
; i
++ ) ans
. m
[ i
] [ i
] = 1 ; while ( power
) { if ( power
& 1 ) ans
= mul ( ans
, base
, n
) ; base
= mul ( base
, base
, n
) ; power
>>= 1 ; } return ans
;
}
void extend_gcd ( ll a
, ll b
, ll
& x
, ll
& y
) { if ( b
== 0 ) { x
= 1 ; y
= 0 ; return ; } extend_gcd ( b
, a
% b
, x
, y
) ; ll tmp
= x
; x
= y
; y
= tmp
- a
/ b
* y
;
}
ll
inverse_mod ( ll y
, ll p
) { ll x
, m
; extend_gcd ( y
, p
, x
, m
) ; return ( p
+ x
% p
) % p
;
}
int degree
[ MAXN
] ;
ll inv
[ MAXN
] ;
int main ( ) { int t
, n
, m
, k
, u
, v
, w
; t
= read ( ) ; while ( t
-- ) { n
= read ( ) ; m
= read ( ) ; k
= read ( ) ; Matrix edge
, mp
; for ( int i
= 0 ; i
< m
; i
++ ) { u
= read ( ) ; v
= read ( ) ; w
= read ( ) ; edge
. m
[ u
] [ v
] = w
+ 1 ; edge
. m
[ v
] [ u
] = w
+ 1 ; } for ( int i
= 1 ; i
<= n
; i
++ ) { for ( int j
= 1 ; j
<= n
; j
++ ) { if ( edge
. m
[ i
] [ j
] == 2 ) { ++ degree
[ i
] ; ++ degree
[ j
] ; ++ degree
[ i
+ n
] ; ++ degree
[ j
+ n
] ; } else if ( edge
. m
[ i
] [ j
] == 1 ) { ++ degree
[ i
] ; ++ degree
[ j
] ; ++ degree
[ i
+ n
] ; ++ degree
[ j
+ n
] ; } } } for ( int i
= 1 ; i
<= n
; i
++ ) { degree
[ i
] >>= 1 ; inv
[ i
] = inverse_mod ( degree
[ i
] , MOD
) ; degree
[ i
] = degree
[ i
+ n
] = 0 ; } for ( int i
= 1 ; i
<= n
; i
++ ) { for ( int j
= 1 ; j
<= n
; j
++ ) { if ( edge
. m
[ i
] [ j
] == 1 ) { mp
. m
[ i
] [ j
] = inv
[ i
] ; mp
. m
[ i
+ n
] [ j
+ n
] = inv
[ i
] ; } else if ( edge
. m
[ i
] [ j
] == 2 ) { mp
. m
[ i
+ n
] [ j
] = inv
[ i
] ; mp
. m
[ i
] [ j
+ n
] = inv
[ i
] ; } } } mp
= fastpow ( mp
, k
, 2 * n
) ; cout
<< mp
. m
[ 1 ] [ 2 * n
] << '\n' ; } return 0 ;
}
樣例沒(méi)問(wèn)題不過(guò)交在HDU上面返回TLE，標(biāo)程就是這個(gè)思路，可能是評(píng)測(cè)機(jī)的問(wèn)題吧，我用榜11 1 500ms500ms 5 0 0 m s 1003鴿子
設(shè)dp[i]dp[i] d p [ i ] ii i ii i jj j dp[i]=min(dp[i]+1,dp[j])dp[i]=min(dp[i]+1,dp[j]) d p [ i ] = m i n ( d p [ i ] + 1 , d p [ j ] ) dp[j]=min(dp[j]+1,dp[i])dp[j]=min(dp[j]+1,dp[i]) d p [ j ] = m i n ( d p [ j ] + 1 , d p [ i ] ) 很容易理解，注意更新不要覆蓋 # include <iostream> # include <cstring> # include <algorithm> # include <queue> # include <stack> # include <vector> # include <cmath> # include <cstdio> # include <map> using namespace std
;
typedef long long ll
;
const int MAXN
= 2e5 + 100 ;
const int INF
= 0x3f3f3f3f ;
int Data
[ MAXN
] ;
int dp
[ MAXN
] ;
int main ( ) { ios
:: sync_with_stdio ( false ) ; int t
, n
, m
, k
, u
, v
; cin
>> t
; while ( t
-- ) { cin
>> n
>> m
>> k
; memset ( dp
, 0x3f , sizeof dp
) ; dp
[ k
] = 0 ; for ( int i
= 1 ; i
<= m
; i
++ ) { cin
>> u
>> v
; int a
= dp
[ u
] ; int b
= dp
[ v
] ; dp
[ v
] = min ( b
+ 1 , a
) ; dp
[ u
] = min ( a
+ 1 , b
) ; } for ( int i
= 1 ; i
<= n
; i
++ ) { if ( i
!= 1 ) cout
<< ' ' ; cout
<< ( dp
[ i
] == INF
? - 1 : dp
[ i
] ) ; } cout
<< '\n' ; } return 0 ;
}
1004萌新
已知aa a bb b cc c aa a bb b cc c c?1>0c-1\gt0 c ? 1 > 0 c≤max(a,b)c\leq max(a,b) c ≤ m a x ( a , b ) 現(xiàn)在假定a>ba\gt b a > b a?b=kca-b=kc a ? b = k c kckc k c a?ba-b a ? b a?b=1a-b=1 a ? b = 1 a?b=0a-b=0 a ? b = 0 a?ba-b a ? b a=1,b=1a=1,b=1 a = 1 , b = 1 # include <iostream> # include <cstring> # include <algorithm> # include <queue> # include <stack> # include <vector> # include <cmath> # include <cstdio> # include <map> using namespace std
;
typedef long long ll
;
const int MAXN
= 2e5 + 100 ;
const int INF
= 0x3f3f3f3f ;
int Data
[ MAXN
] ;
int main ( ) { ios
:: sync_with_stdio ( false ) ; int t
; ll a
, b
; cin
>> t
; while ( t
-- ) { cin
>> a
>> b
; if ( a
< b
) swap ( a
, b
) ; ll c
= a
- b
; if ( ( a
== 1 && b
== 1 ) || c
== 1 ) { cout
<< - 1 << " " << - 1 << '\n' ; continue ; } if ( a
== b
) { cout
<< 2 << " " << a
<< '\n' ; continue ; } int f
= 0 ; for ( ll i
= 2 ; i
* i
<= c
; i
++ ) { if ( c
% i
== 0 ) { cout
<< i
<< " " ; f
= 1 ; break ; } } if ( f
) cout
<< c
<< '\n' ; else cout
<< c
<< ' ' << c
<< '\n' ; } return 0 ;
}
1006毒瘤數(shù)據(jù)結(jié)構(gòu)題
考察思維，用兩個(gè)指針維護(hù)最左邊的兩個(gè)0，對(duì)于操作1，如果修改了第一個(gè)0，則更新最左邊兩個(gè)0的位置，如果修改了第二個(gè)0，則更新第二個(gè)0的位置，其他情況正常標(biāo)記；對(duì)于操作2，如果查詢到第一個(gè)0，那么答案為第二個(gè)0的位置，其他情況答案都為第一個(gè)0的位置 在HDU上必須用超級(jí)快讀才能過(guò)得了 # include <iostream> # include <algorithm> # include <cstring> # include <cstdio> # include <vector> # include <cmath> # include <queue> # include <stack> # include <map> # include <iomanip> using namespace std
;
typedef long long ll
;
typedef unsigned long long ull
;
const int INF
= 0x3f3f3f3f ;
const int MAXN
= 1e7 + 100 ;
const double eps
= 1e-6 ;
int vis
[ MAXN
] ;
inline int read ( ) { int x
= 0 , f
= 1 ; char c
= getchar ( ) ; while ( c
< '0' || c
> '9' ) { if ( c
== '-' ) f
= - 1 ; c
= getchar ( ) ; } while ( c
>= '0' && c
<= '9' ) { x
= ( x
<< 1 ) + ( x
<< 3 ) + ( c
^ 48 ) ; c
= getchar ( ) ; } return x
* f
;
}
namespace IO
{ inline char nc ( ) { static char buf
[ 100000 ] , * p1
= buf
, * p2
= buf
; return p1
== p2
&& ( p2
= ( p1
= buf
) + fread ( buf
, 1 , 100000 , stdin ) , p1
== p2
) ? EOF : * p1
++ ; } template < typename T > inline T
read ( ) { char ch
= nc ( ) ; T sum
= 0 ; while ( ch
< '0' || ch
> '9' ) { ch
= nc ( ) ; if ( ch
== EOF ) return EOF ; } while ( ch
>= '0' && ch
<= '9' ) { sum
= ( sum
<< 1 ) + ( sum
<< 3 ) + ( ch
^ 48 ) ; ch
= nc ( ) ; if ( ch
== EOF ) return EOF ; } return sum
; }
}
int main ( ) { ios
:: sync_with_stdio ( false ) ; int n
, a
, b
; n
= IO
:: read
< int > ( ) ; int x
, y
; x
= 1 ; y
= 2 ; for ( int i
= 0 ; i
< n
; i
++ ) { a
= IO
:: read
< int > ( ) ; b
= IO
:: read
< int > ( ) ; if ( a
== 1 ) { vis
[ b
] = 1 ; if ( b
== y
) { while ( vis
[ y
] ) ++ y
; } else if ( b
== x
) { x
= y
; ++ y
; while ( vis
[ y
] ) ++ y
; } } else { if ( b
== x
) printf ( "%d\n" , y
) ; else printf ( "%d\n" , x
) ; } } return 0 ;
}
1008獵人殺
狼人先殺人（也可能自刀），之后獵人開槍互相殺，直到殺死狼或者剩下人數(shù)不大于2 暴力模擬即可 # include <iostream> # include <algorithm> # include <cstring> # include <cstdio> # include <vector> # include <cmath> # include <queue> # include <stack> # include <map> # include <iomanip> using namespace std
;
typedef long long ll
;
typedef unsigned long long ull
;
const int INF
= 0x3f3f3f3f ;
const int MAXN
= 1e6 + 100 ;
const double eps
= 1e-6 ;
int main ( ) { ios
:: sync_with_stdio ( false ) ; cin
. tie ( 0 ) ; cout
. tie ( 0 ) ; int t
, n
, x
; cin
>> t
; while ( t
-- ) { cin
>> n
; int id
= - 1 ; for ( int i
= 1 ; i
<= n
; i
++ ) { cin
>> x
; if ( x
== 1 ) id
= i
; } vector
< vector
< int > > vs ( n
+ 5 ) ; vector
< int > vis ( n
+ 5 ) ; for ( int i
= 1 ; i
<= n
; i
++ ) { for ( int j
= 1 ; j
<= n
; j
++ ) { cin
>> x
; vs
[ i
] . push_back ( x
) ; } } int ok
= 0 ; int now
= vs
[ id
] [ 0 ] ; int res
= n
- 1 ; vis
[ now
] = 1 ; if ( now
== id
) { ok
= 2 ; } else if ( res
<= 2 ) { ok
= 1 ; } else { while ( 1 ) { if ( ok
) break ; for ( int i
= 0 ; i
< vs
[ now
] . size ( ) ; i
++ ) { int to
= vs
[ now
] [ i
] ; if ( vis
[ to
] || to
== now
) continue ; vis
[ to
] = 1 ; now
= to
; -- res
; if ( now
== id
) { ok
= 2 ; } else if ( res
<= 2 ) { ok
= 1 ; } break ; } } } cout
<< ( ok
== 1 ? "langren" : "lieren" ) << '\n' ; } return 0 ;
}
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