【codeforces】Codeforces Round #277 (Div. 2) 解讀

門戶：Codeforces Round #277 (Div. 2)

486A. Calculating Function

裸公式= =

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

typedef long long LL ;

LL n ;

int main () {
	while ( ~scanf ( "%I64d" , &n ) ) printf ( "%I64d\n" , n / 2 - ( n % 2 ?

n : 0 ) ) ; return 0 ; }

486B. OR in Matrix

依據題意，将a矩陣必須為0的地方先填充為0，其它地方置為1，然後對b矩陣中為1的bij推斷第i行或第j列是否有1，沒有輸出NO，推斷到最後假設全部的bij都是合法的，則輸出YES。

#include <map>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define mid ( ( l + r ) >> 1 )

int a[105][105] ;
int b[105][105] ;
int R[105] , C[105] ;
int n , m ;

void solve () {
	clr ( R , 0 ) ;
	clr ( C , 0 ) ;
	For ( i , 1 , n ) For ( j , 1 , m ) a[i][j] = 1 ;
	For ( i , 1 , n ) For ( j , 1 , m ) {
		scanf ( "%d" , &b[i][j] ) ;
		if ( !b[i][j] ) {
			For ( k , 1 , m ) a[i][k] = 0 ;
			For ( k , 1 , n ) a[k][j] = 0 ;
		}
	}
	For ( i , 1 , n ) For ( j , 1 , m ) {
		R[i] += a[i][j] ;
		C[j] += a[i][j] ;
	}
	For ( i , 1 , n ) For ( j , 1 , m ) if ( b[i][j] ) {
		if ( a[i][j] ) continue ;
		if ( R[i] || C[j] ) continue ;
		printf ( "NO\n" ) ;
		return ;
	}
	printf ( "YES\n" ) ;
	For ( i , 1 , n ) For ( j , 1 , m ) printf ( "%d%c" , a[i][j] , j < m ?

' ' : '\n' ) ; } int main () { while ( ~scanf ( "%d%d" , &n , &m ) ) solve () ; return 0 ; }

486C. Palindrome Transformation

貪心，其實僅僅用在初始位置所在的字元串半邊處理便足夠了，于是考慮幾種情況，判一下就可以。這題錯的吐血了。少打兩個else，導緻絕殺失敗，不然30多名好歹好看一點。

。

#include <map>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define mid ( ( l + r ) >> 1 )

const int MAXN = 100005 ;

char s[MAXN] ;
int a[MAXN] ;
int n , p ;

void solve () {
	int ans = 0 ;
	clr ( a , 0 ) ;
	scanf ( "%s" , s + 1 ) ;
	int l = n + 1 , r = 1 ;
	For ( i , 1 , n / 2 ) {
		if ( s[i] != s[n - i + 1] ) {
			a[i] = abs ( s[i] - s[n - i + 1] ) ;
			a[n - i + 1] = a[i] = min ( a[i] , 26 - a[i] ) ;
			ans += a[i] ;
		}
	}
	if ( p <= n / 2 ) {
		For ( i , 1 , n / 2 ) if ( a[i] ) l = min ( l , i ) , r = max ( r , i ) ;
	} else {
		For ( i , n / 2 + 1 , n ) if ( a[i] ) l = min ( l , i ) , r = max ( r , i ) ;
	}
	if ( l != n + 1 ) {
		if ( p <= l ) ans += r - p ;
		else if ( p >= r ) ans += p - l ;
		else ans += min ( r - l + r - p , r - l + p - l ) ;
	}
	printf ( "%d\n" , ans ) ;
}

int main () {
	while ( ~scanf ( "%d%d" , &n , &p ) ) solve () ;
	return 0 ;
}

486D. Valid Sets

考慮以每一個點作為根結點擴充出一棵樹，這個樹滿足樹上全部的節點的權值都不比樹根大且val[root]-val[v]<=d。然後能夠樹型DP求以這個點為樹根的集合數。考慮到假設以u為根時擴充的樹中包括了與u權值同樣的v，那麼以v為根時便不能包括u了，這個我們能夠用一個數組判重。

詳細見代碼。

#include <map>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define mid ( ( l + r ) >> 1 )

const int MAXN = 2005 ;
const int MAXE = 4005 ;
const int mod = 1e9 + 7 ;

struct Edge {
    int v , n ;
    Edge () {}
    Edge ( int v , int n ) : v ( v ) , n ( n ) {}
} ;

Edge E[MAXE] ;
int H[MAXN] , cntE ;
int val[MAXN] ;
int vis[MAXN][MAXN] ;
LL dp[MAXN] ;
int d , n ;
int root ;

void clear () {
    cntE = 0 ;
    clr ( H , -1 ) ;
    clr ( vis , 0 ) ;
}

void addedge ( int u , int v ) {
    E[cntE] = Edge ( v , H[u] ) ;
    H[u] = cntE ++ ;
}

LL dfs ( int u , int fa ) {
    dp[u] = 1 ;
    LL ans = 1 ;
    for ( int i = H[u] ; ~i ; i = E[i].n ) {
        int v = E[i].v ;
        if ( v == fa ) continue ;
        if ( val[v] > val[root] || vis[root][v] || val[root] - val[v] > d ) continue ;
        if ( val[root] == val[v] ) vis[root][v] = vis[v][root] = 1 ;
        LL tmp = dfs ( v , u ) ;
        ans = ( ans + tmp * dp[u] % mod ) % mod ;
        dp[u] = ( dp[u] + tmp * dp[u] ) % mod ;
    }
    return ans ;
}

void solve () {
    int u , v ;
    clear () ;
    For ( i , 1 , n ) scanf ( "%d" , &val[i] ) ;
    rep ( i , 1 , n ) {
        scanf ( "%d%d" , &u , &v ) ;
        addedge ( u , v ) ;
        addedge ( v , u ) ;
    }
    LL ans = 0 ;
    For ( i , 1 , n ) {
        root = i ;
        ans = ( ans + dfs ( i , 0 ) ) % mod ;
    }
    printf ( "%I64d\n" , ans ) ;
}

int main () {
    while ( ~scanf ( "%d%d" , &d , &n ) ) solve () ;
    return 0 ;
}

486E. LIS of Sequence

設F1[i]為1~i内以i結尾的LIS。F2[i]為i~n内以i開頭的LIS。ans為1~n内的LIS。

1.F1[i]+F2[i]-1<ans。

2.F1[i]+F2[i]-1==ans時長度F1[i]不唯一。

3.F1[i]+F2[i]-1==ans時長度F1[i]唯一。

可用二分求F1[i]，F2[i]。

#include <map>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = ( a ) ; i <  ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define mid ( ( l + r ) >> 1 )

const int MAXN = 100005 ;

int vis[MAXN] ;
int S[MAXN] , top ;
int a[MAXN] ;
int F1[MAXN] ;
int F2[MAXN] ;
int n ;

int search1 ( int x , int l , int r ) {
	while ( l < r ) {
		int m = mid ;
		if ( S[m] >= x ) r = m ;
		else l = m + 1 ;
	}
	return l ;
}

int search2 ( int x , int l , int r ) {
	while ( l < r ) {
		int m = mid ;
		if ( S[m] <= x ) r = m ;
		else l = m + 1 ;
	}
	return l ;
}

void solve () {
	clr ( vis , 0 ) ;
	For ( i , 1 , n ) scanf ( "%d" , &a[i] ) ;
	top = 0 ;
	For ( i , 1 , n ) {
		if ( !top || a[i] > S[top] ) {
			S[++ top] = a[i] ;
			F1[i] = top ;
		} else {
			int x = search1 ( a[i] , 1 , top ) ;
			S[x] = a[i] ;
			F1[i] = x ;
		}
	}
	top = 0 ;
	rev ( i , n , 1 ) {
		if ( !top || a[i] < S[top] ) {
			S[++ top] = a[i] ;
			F2[i] = top ;
		} else {
			int x = search2 ( a[i] , 1 , top ) ;
			S[x] = a[i] ;
			F2[i] = x ;
		}
	}
	int ans = top ;
	For ( i , 1 , n ) if ( F1[i] + F2[i] - 1 == ans ) ++ vis[F1[i]] ;
	For ( i , 1 , n ) {
		if ( F1[i] + F2[i] - 1 < ans ) putchar ( '1' ) ;
		else if ( vis[F1[i]] == 1 ) putchar ( '3' ) ;
		else putchar ( '2' ) ;
	}
	printf ( "\n" ) ;
}

int main () {
	while ( ~scanf ( "%d" , &n ) ) solve () ;
	return 0 ;
}