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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....
2015.7.17( NOI2015 day1 )
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原文地址：http://www.cnblogs.com/JSZX11556/p/4654928.html
