洛谷 P1309 瑞士輪 歸併

2021-10-23 20:52:34 字數 2386 閱讀 7488

每一輪快排,超時:

# include

# include

# include

# include

using

namespace std;

class

athlete

// athlete

(int _id,

int _competence,

int _score):id

(_id)

,competence

(_competence)

,score

(_score)};

bool

comp

(const athlete& _a,

const athlete& _b)

intmain()

sort

(athlete_vec.

begin()

,athlete_vec.

end(

),comp)

;for

(int round=

0;round++round)

sort

(athlete_vec.

begin()

,athlete_vec.

end(

),comp);}

cout<.id+

1

init_score_arr;

delete

competence_arr;

return0;

}

每次比賽之後,分數按照降序排列,每兩人中的贏的人都+1分,贏的n個人仍為降序,輸的n個人也仍為降序,所以將贏的序列與輸的序列歸併即可。每輪o(n)複雜度。而快排每輪o(nlogn)。

用vector push_back出現超時,用new分配空間accepeted,應該是push_back太慢。

# include

# include

# include

# include

using

namespace std;

class

athlete

// athlete

(int _id,

int _competence,

int _score):id

(_id)

,competence

(_competence)

,score

(_score)};

bool

comp

(const athlete& _a,

const athlete& _b)

intmain()

sort

(athlete_vec.

begin()

,athlete_vec.

end(

),comp)

;for

(int i=

0;i<

2*n;

++i)

int* winner_id_arr=

newint

[n];

int* loser_id_arr=

newint

[n];

for(

int round=

0;round++round)

else

winner_id_arr[i]

=winner_id;

loser_id_arr[i]

=loser_id;

}// printf("winner_id_vec: ");

// for(int i=0;i// printf("%d, ",winner_id_vec[i]);

// cout//merge sort

int winner_i=0;

int loser_i=0;

int i=0;

while

(winner_ielse

}while

(winner_iwhile

(loser_i// printf("rank: ");

// for(int i=0;i<2*n;++i)

// printf("%d, ",rank_id_arr[i]);

// cout<}

cout<+1

winner_id_arr;

delete

loser_id_arr;

delete

score_arr;

delete

competence_arr;

delete

rank_id_arr;

return0;

}

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