2015.7.17( NOI2015 day1 )

今天早起做了NOI2015网络同步赛....

最近NOI是越来越向NOIP靠拢了....但是我还是不会做.....

第一题:程序自动分析

先离散化一下..然后最多就剩20w个数 , 不等于就存起来..等于就用并查集维护...最后for判断一下就行了.....

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<cctype>

#include<cstdlib>

 

#define rep(i ,n) for(int i=0; i < n; ++i)

#define clr(x ,c) memset(x, c, sizeof(x))

 

using namespace std;

 

const int maxn = 200005;

 

struct T {

int x;

T* next;

} *head[ maxn ] , pool[ maxn ] , *pt;

 

void add( int u , int v ) {

pt -> x = v;

pt -> next = head[ u ];

head[ u ] = pt++;

}

 

inline int read() {

char c = getchar();

int ans = 0;

for( ; ! isdigit( c ) ; c = getchar() );

for( ; isdigit( c ) ; c = getchar() )

   ans = ans * 10 + c - '0';

return ans;

}

 

int fa[ maxn ];

 

int find( int x ) {

return x == fa[ x ] ? x : fa[ x ] = find( fa[ x ] );

}

 

struct Q {

int x , y;

bool type;

} q[ maxn ];

 

int id[ maxn << 1 ];

 

int main(){

freopen("prog.in", "r", stdin);

freopen("prog.out", "w", stdout);

int t = read();

while( t-- ) {

int cnt = 0;

pt = pool;

int n = read();

rep( i , n ) {

Q &p = q[ i ];

   id[ cnt++ ] = p.x = read();

id[ cnt++ ] = p.y = read();

p.type = read();

}

sort( id , id + cnt );

cnt = unique( id , id + cnt ) - id;

rep( i , cnt ) fa[ i ] = i;

clr( head , 0 );

rep( i , n ) {

Q* p = q + i;

if( p -> type ) 

   fa[ find( lower_bound( id , id + cnt , p -> x ) - id ) ] = 

       find( lower_bound( id , id + cnt , p -> y ) - id );

else 

   add( lower_bound( id , id + cnt , p -> x ) - id ,

        lower_bound( id , id + cnt , p -> y ) - id );

}

bool ans = true;

rep( i , cnt ) {

for( T* p = head[ i ] ; p ; p = p -> next ) {

   if( find( i ) == find( p -> x ) ) {

    ans = false;

    break;

   }

}

if( ! ans ) break;

}

printf( ans ? "YES\n" : "NO\n" );

}

return 0;

} 

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第二题 : 软件包管理器

树链剖分... install( x ) 就处理 x 到 root 的这条路径 , uninstall( x ) 就处理以 x 为根的子树...子树在线段树上是连续的.. 两个操作都可以打标...话说我对拍时暴力程序也跑得很快...难道暴力也能过 ?

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<cctype>

#include<cstdlib>

 

#define rep(i ,n) for(int i=0; i < n; ++i)

#define clr(x ,c) memset(x, c, sizeof(x))

#define REP( x ) for( edge* e = head[ x ] ; e ; e = e -> next )

#define M( l , r ) ( ( ( l ) + ( r ) ) >> 1 )

 

using namespace std;

 

const int maxn = 100000 + 5;

 

struct edge {

int to;

edge* next;

} E[ maxn << 1 ] , *ptt = E , *head[ maxn ];

 

inline void add( int u , int v ) {

ptt -> to = v;

ptt -> next = head[ u ];

head[ u ] = ptt++;

}

#define add_edge( u , v ) add( u , v ) , add( v , u )

 

int top[ maxn ] , fa[ maxn ] , dep[ maxn ] , size[ maxn ] , son[ maxn ];

int id[ maxn ] , id_cnt = 0 , TOP , n , _L[ maxn ] , _R[ maxn ];

 

void dfs( int x ) {

size[ x ] = 1 , son[ x ] = -1;

REP( x ) {

int to = e -> to;

if( to == fa[ x ] ) continue;

dep[ to ] = dep[ x ] + 1;

fa[ to ] = x;

dfs( to );

size[ x ] += size[ to ];

if( son[ x ] == -1 || size[ son[ x ] ] < size[ to ] ) 

  son[ x ] = to;

}

}

 

void DFS( int x ) {

top[ x ] = TOP;

_L[ x ] = id[ x ] = ++id_cnt;

if( son[ x ] != -1 ) DFS( son[ x ] );

REP( x ) if( fa[ x ] != e -> to && e -> to != son[ x ] )

   DFS( TOP = e -> to );

_R[ x ] = id_cnt;

}

 

void init() {

dfs( dep[ 0 ] = 0 );

DFS( TOP = 0 );

}

 

struct Node {

Node *l , *r;

int set , sum; // sum : sum of install

Node() {

set = sum = 0;

l = r = NULL;

}

inline void update( int len ) {

if( set != -1 )

   sum = set * len;

else 

   sum = l -> sum + r -> sum;

}

inline void pushdown() {

if( set != -1 ) {

l -> set = r -> set = set;

set = -1;

}

}

} pool[ maxn << 1 ] , *pt = pool , *root;

 

void build( Node* t , int l , int r ) {

if( r > l ) {

int m = M( l , r );

build( t -> l = pt++ , l , m );

build( t -> r = pt++ , m + 1 , r );

}

}

 

int L , R;

bool v;// v -> true install

 

void modify( Node* t , int l , int r ) {

if( L <= l && r <= R ) {

t -> set = v;

} else {

int m = M( l , r );

t -> pushdown();

L <= m ? modify( t -> l , l , m ) : t -> l -> update( m - l + 1 );

m < R ? modify( t -> r , m + 1 , r ) : t -> r -> update( r - m );

}

t -> update( r - l + 1 );

}

 

int query( Node* t , int l , int r ) {

if( L <= l && r <= R )

return v ? r - l + 1 - t -> sum : t -> sum;

int m = M( l , r );

t -> pushdown();

t -> l -> update( m - l + 1 );

t -> r -> update( r - m );

return ( L <= m ? query( t -> l , l , m ) : 0 ) + 

      ( m < R ? query( t -> r , m + 1 , r ) : 0 );

}

 

int install( int x , int y ) {

int ans = 0;

while( top[ x ] != top[ y ] ) {

if( dep[ top[ x ] ] < dep[ top[ y ] ] ) swap( x , y );

L = id[ top[ x ] ] , R = id[ x ];

ans += query( root , 1 , n );

modify( root , 1 , n );

x = fa[ top[ x ] ];

}

if( dep[ x ] < dep[ y ] ) swap( x , y );

L = id[ y ] , R = id[ x ];

ans += query( root , 1 , n );

modify( root , 1 , n );

return ans;

}

 

inline int read() {

char c = getchar();

int ans = 0;

for( ; ! isdigit( c ) ; c = getchar() );

for( ; isdigit( c ) ; c = getchar() )

   ans = ans * 10 + c - '0';

return ans;

}

 

int main(){

freopen("manager.in", "r", stdin);

freopen("manager.out", "w", stdout);

clr( head , 0 );

n = read();

build( root = pt++ , 1 , n );

rep( i , n - 1 ) {

int t = read();

   add_edge( t , i + 1 );

}

init();

int q = read() , x;

char s[ 10 ];

while( q-- ) {

scanf( "%s%d" , s , &x );

if( s[ 0 ] == 'i' ) {

v = true;

printf( "%d\n" , install( 0 , x ) );

} else {

v = false;

L = _L[ x ] , R = _R[ x ];

printf( "%d\n" , query( root , 1 , n ) );

modify( root , 1 , n );

}

}

return 0;

} 

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第三题 : 寿司晚宴

完全不会..但是打表还能弄点分吧....暴力程序打个表然后就30分了...

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<cctype>

#include<cstdlib>

 

#define rep(i ,n) for(int i=0; i < n; ++i)

#define clr(x ,c) memset(x, c, sizeof(x))

 

using namespace std;

 

typedef long long ll;

 

const int maxn = 50;

ll ans[ maxn ];

 

int main(){

freopen("dinner.in", "r", stdin);

freopen("dinner.out", "w", stdout);

ans[ 2 ] = 3;

    ans[ 3 ] = 9;

    ans[ 4 ] = 21;

    ans[ 5 ] = 63;

    ans[ 6 ] = 111;

    ans[ 7 ] = 333;

    ans[ 8 ] = 693;

    ans[ 9 ] = 1521;

    ans[ 10 ] = 2577;

    ans[ 11 ] = 7731;

    ans[ 12 ] = 13491;

    ans[ 13 ] = 40473;

    ans[ 14 ] = 67833;

    ans[ 15 ] = 119241;

    ans[ 16 ] = 239481;

    ans[ 17 ] = 718443;

    ans[ 18 ] = 1340523;

    ans[ 19 ] = 4021569;

    ans[ 20 ] = 7494849;

    ans[ 21 ] = 13356657;

    ans[ 22 ] = 22271409;

    ans[ 23 ] = 66814227;

    ans[ 24 ] = 130266387;

    ans[ 25 ] = 268286823;

    ans[ 26 ] = 447212583;

    ans[ 27 ] = 896472063;

    ans[ 28 ] = 1684872063;

    ans[ 29 ] = 5054616189;

ll n , MOD;

cin >> n >> MOD;

    if( n < 30 ) cout << ans[ n ] % MOD << "\n";

    else cout << 1LL * rand() * rand() % MOD << "\n";

return 0;

} 

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后天还有day2...没玩过提答题 , 挺想玩的....