【ACdream】Andrew Stankevich Contest （1）

A：1210 Chinese Girls' Amusement

大数。如果n%4==0，输出n/2-1，如果n%2==0&&n%4==1，输出n/2-2，如果n%2==1，输出n/2。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>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 travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )#define clr( a , x ) memset ( a , x , sizeof a )#define cpy( a , x ) memcpy ( a , x , sizeof a )#define ls ( o << 1 )#define rs ( o << 1 | 1 )#define lson ls , l , m#define rson rs , m + 1 , r#define mid ( ( l + r ) >> 1 )#define root 1 , 1 , nconst int MAXN = 100006 ; char s[MAXN] ;int num[MAXN] ; void solve () {int n = strlen ( s ) ;rep ( i , 0 , n ) num[i] = s[i] - '0' ;int flag = num[n - 1] & 1 ;int tmp = 0 ;rep ( i , 0 , n ) {num[i] += tmp * 10 ;tmp = num[i] % 2 ;num[i] /= 2 ;}if ( flag == 1 ) {if ( num[0] ) printf ( "%d" , num[0] ) ;rep ( i , 1 , n ) printf ( "%d" , num[i] ) ;printf ( "\n" ) ;} else {int del = 1 ;rev ( i , n - 1 , 0 ) {del = num[i] - del ;num[i] = del ;del = 0 ;if ( num[i] < 0 ) {num[i] += 10 ;del = 1 ;}}if ( ( num[n - 1] & 1 ) == 0 ) {int del = 1 ;rev ( i , n - 1 , 0 ) {del = num[i] - del ;num[i] = del ;del = 0 ;if ( num[i] < 0 ) {num[i] += 10 ;del = 1 ;}}}if ( num[0] ) printf ( "%d" , num[0] ) ;rep ( i , 1 , n ) printf ( "%d" , num[i] ) ;printf ( "\n" ) ;}} int main () {while ( gets ( s ) ) solve () ;return 0 ;}

B：1211 Reactor Cooling

裸的带下界的可行流。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>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 travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )#define clr( a , x ) memset ( a , x , sizeof a )#define cpy( a , x ) memcpy ( a , x , sizeof a )#define ls ( o << 1 )#define rs ( o << 1 | 1 )#define lson ls , l , m#define rson rs , m + 1 , r#define mid ( ( l + r ) >> 1 )#define root 1 , 1 , n const int MAXN = 205 ;const int MAXE = 1000000 ;const int INF = 0x3f3f3f3f ; struct Edge {int v , c , n ;Edge () {}Edge ( int v , int c , int n ) : v ( v ) , c ( c ) , n ( n ) {}} E[MAXE] ; int H[MAXN] , cntE ;int d[MAXN] , cur[MAXN] , pre[MAXN] , num[MAXN] ;int Q[MAXN] , head , tail ;int s , t , nv ;int flow ;int n , m ;int in[MAXN] , ou[MAXN] ;int low[MAXE] ;void clear () {cntE = 0 ;clr ( H , -1 ) ;clr ( in , 0 ) ;clr ( ou , 0 ) ;} void addedge ( int u , int v , int c ) {E[cntE] = Edge ( v , c , H[u] ) ;H[u] = cntE ++ ;E[cntE] = Edge ( u , 0 , H[v] ) ;H[v] = cntE ++ ;} void rev_bfs () {clr ( d , -1 ) ;clr ( num , 0 ) ;head = tail = 0 ;Q[tail ++] = t ;d[t] = 0 ;num[0] = 1 ;while ( head != tail ) {int u = Q[head ++] ;for ( int i = H[u] ; ~i ; i = E[i].n ) {int v = E[i].v ;if ( ~d[v] ) continue ;d[v] = d[u] + 1 ;num[d[v]] ++ ;Q[tail ++] = v ;}}} int ISAP () {rev_bfs () ;cpy ( cur , H ) ;flow = 0 ;int u = pre[s] = s , i , pos , f , minv ;while ( d[s] < nv ) {if ( u == t ) {f = INF ;for ( i = s ; i != t ; i = E[cur[i]].v ) {if ( f > E[cur[i]].c ) {f = E[cur[i]].c ;pos = i ;}}for ( i = s ; i != t ; i = E[cur[i]].v ) {E[cur[i]].c -= f ;E[cur[i] ^ 1].c += f ;}flow += f ;u = pos ;}for ( i = cur[u] ; ~i ; i = E[i].n ) {if ( E[i].c && d[u] == d[E[i].v] + 1 ) break ;}if ( ~i ) {cur[u] = i ;pre[E[i].v] = u ;u = E[i].v ;} else {if ( 0 == --num[d[u]] ) break ;for ( minv = nv , i = H[u] ; ~i ; i = E[i].n ) {if ( E[i].c && minv > d[E[i].v] ) {minv = d[E[i].v] ;cur[u] = i ;}}d[u] = minv + 1 ;num[d[u]] ++ ;u = pre[u] ;}}return flow ;} void solve () {int u , v , c ;clear () ;s = 0 ;t = n + 1 ;nv = t + 1 ;rep ( i , 0 , m ) {scanf ( "%d%d%d%d" , &u , &v , &low[i] , &c ) ;in[v] += low[i] ;ou[u] += low[i] ;addedge ( u , v , c - low[i] ) ;}For ( i , 1 , n ) {int tmp = in[i] - ou[i] ;if ( tmp > 0 ) addedge ( s , i ,  tmp ) ;if ( tmp < 0 ) addedge ( i , t , -tmp ) ;}ISAP () ;int ok = 1 ;for ( int i = H[s] ; ~i ; i = E[i].n ) if ( E[i].c ) ok = 0 ;if ( ok == 0 ) {printf ( "NO\n" ) ;} else {printf ( "YES\n" ) ;rep ( i , 0 , m ) printf ( "%d\n" , low[i] + E[i << 1 | 1].c ) ;}} int main () {while ( ~scanf ( "%d%d" , &n , &m ) ) solve () ;return 0 ;}

D：1213 Matrix Multiplication

所有边的两个点的度之和即答案。观察找规律可知，两个矩阵相乘时如果对应位置为1则必定是两条线段有公共点。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>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 travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )#define clr( a , x ) memset ( a , x , sizeof a )#define cpy( a , x ) memcpy ( a , x , sizeof a )#define ls ( o << 1 )#define rs ( o << 1 | 1 )#define lson ls , l , m#define rson rs , m + 1 , r#define mid ( ( l + r ) >> 1 )#define root 1 , 1 , nconst int MAXN = 10005 ;const int MAXE = 100005 ; struct Seg {int u , v ;} seg[MAXE] ; int num[MAXN] ;int n , m ; void solve () {clr ( num , 0 ) ;rep ( i , 0 , m ) {scanf ( "%d%d" , &seg[i].u , &seg[i].v ) ;num[seg[i].u] ++ ;num[seg[i].v] ++ ;}LL ans = 0 ;rep ( i , 0 , m ) {ans += num[seg[i].u] ;ans += num[seg[i].v] ;}printf ( "%lld\n" , ans ) ;} int main () {while ( ~scanf ( "%d%d" , &n , &m ) ) solve () ;return 0 ;}

E：1214 Nice Patterns Strike Back

因为下一列只与上一列的状态有关，可以暴力判断预处理出转移矩阵。高精度+矩阵快速幂，比赛的时候高精度写错了，蛋疼死。。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>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 travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )#define clr( a , x ) memset ( a , x , sizeof a )#define cpy( a , x ) memcpy ( a , x , sizeof a )#define ls ( o << 1 )#define rs ( o << 1 | 1 )#define lson ls , l , m#define rson rs , m + 1 , r#define mid ( ( l + r ) >> 1 )#define root 1 , 1 , cnt const int M = 50 ;const int L = 10 ; int mod ; struct BigInt {int digit[1005] ;int length ;BigInt () {}BigInt ( char buf[] ) : length ( 0 ) {clr ( digit , 0 ) ;length = strlen ( buf ) ;rep ( i , 0 , length ) digit[i] = buf[length - i - 1] - '0' ;}int operator [] ( const int& index ) const {return digit[index] ;}int& operator [] ( const int& index ) {return digit[index] ;}BigInt fix () {while ( length && digit[length - 1] == 0 ) -- length ;return *this ;}BigInt operator / ( const int& b ) {BigInt c ;c.length = length ;int tmp = 0 ;rev ( i , length - 1 , 0 ) {c[i] = digit[i] + tmp * 10 ;tmp = c[i] % b ;c[i] /= b ;}return c.fix () ;}BigInt operator - ( const int& b ) {BigInt c ;c.length = length ;int tmp = b ;rep ( i , 0 , length ) {tmp = digit[i] - tmp ;c[i] = tmp ;tmp = 0 ;if ( c[i] < 0 ) {c[i] += 10 ;tmp = 1 ;}}return c.fix () ;}void show () {if ( length == 0 ) {printf ( "0\n" ) ;return ;}printf ( "%d" , digit[length - 1] ) ;rev ( i , length - 2 , 0 ) printf ( "%d" , digit[i] ) ;printf ( "\n" ) ;}} ; struct Matrix {int mat[M][M] ;int n ;Matrix () {clr ( mat , 0 ) ;}Matrix ( int n ) : n ( n ) {clr ( mat , 0 ) ;}void eyes () {clr ( mat , 0 ) ;rep ( i , 0 , n ) mat[i][i] = 1 ;}void empty () {clr ( mat , 0 ) ;}Matrix operator * ( const Matrix& a ) const {Matrix c ( n ) ;rep ( i , 0 , n ) rep ( j , 0 , n ) rep ( k , 0 , n ) {c.mat[i][j] = ( c.mat[i][j] + mat[i][k] * a.mat[k][j] ) % mod ;}return c ;}void show () {rep ( i , 0 , n ) {rep ( j , 0 , n ) printf ( "%d " , mat[i][j] ) ;printf ( "\n" ) ;}}} ; Matrix A[6] ;char buf[500] ;int m ; int check ( int a , int b , int n ) {int cnt1 = 0 ;int cnt2 = 0 ;rep ( i , 0 , n ) {if ( a % 2 == 0 && b % 2 == 0 ) {cnt1 ++ ;cnt2 = 0 ;if ( cnt1 == 2 ) return 0 ;} else if ( a % 2 == 1 && b % 2 == 1 ) {cnt2 ++ ;cnt1 = 0 ;if ( cnt2 == 2 ) return 0 ;} else cnt1 = cnt2 = 0 ;a >>= 1 ;b >>= 1 ;}return 1 ;} void fun () {For ( i , 1 , 5 ) {int S = 1 << i ;A[i].n = S ;A[i].empty () ;rep ( j , 0 , S ) {rep ( k , 0 , S ) {if ( check ( j , k , i ) ) A[i].mat[j][k] = 1 ;}}}} Matrix pow ( Matrix& a , BigInt& b ) {Matrix res ( a.n ) ;Matrix tmp = a ;res.eyes () ;while ( b.length ) {if ( b[0] & 1 ) res = res * tmp ;tmp = tmp * tmp ;b = b / 2 ;}return res ;}  void solve () {BigInt c ( buf ) ;c = c - 1 ;Matrix res = pow ( A[m] , c ) ;int S = 1 << m ;int ans = 0 ;rep ( i , 0 , S ) rep ( j , 0 , S ) ans = ( ans + res.mat[i][j] ) % mod ;printf ( "%d\n" , ans ) ;} int main () {fun () ;while ( ~scanf ( "%s%d%d" , buf , &m ,&mod ) ) solve () ;return 0 ;}

G：1216 Beautiful People

线段树优化DP。

先将所有点按照第一个关键字排序，将第二个关键字离散化。然后按第一关键字从小到大扫所有的人，一段一段的更新第一关键字相同的。每一段相同关键字的人用query求出他的最优值，以及最优值对应的id，然后更新前驱，接下来用栈保存他的信息，直到所有第一关键字相同的人都查询完后，再将栈中保存的信息用以更新线段树，线段树中以b作为下标，每个区间保存这个区间的最大单调子序列的长度以及结尾元素的id。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>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 travel( e , H , u ) for ( Edge* e = H[u] ; e ; e = e -> next )#define clr( a , x ) memset ( a , x , sizeof a )#define cpy( a , x ) memcpy ( a , x , sizeof a )#define ls ( o << 1 )#define rs ( o << 1 | 1 )#define lson ls , l , m#define rson rs , m + 1 , r#define mid ( ( l + r ) >> 1 )#define root 1 , 1 , cntconst int MAXN = 300005 ; struct Node {int a , b , idx ;bool operator < ( const Node& t ) const {return a < t.a ;}} node[MAXN] ; struct Stack {int pos , max , idx ;Stack () {}Stack ( int pos , int max , int idx ) : pos ( pos ) , max ( max ) , idx ( idx ) {}} S[MAXN] ; int top ;int pre[MAXN] ;int maxv[MAXN << 2] ;int idx[MAXN << 2] ;int a[MAXN] , cnt ;int n ; int unique ( int n ) {int cnt = 1 ;sort ( a + 1 , a + n + 1 ) ;For ( i , 2 , n ) if ( a[i] != a[cnt] ) a[++ cnt] = a[i] ;return cnt ;} int hash ( int x ) {int l = 1 , r = cnt ;while ( l < r ) {int m = ( l + r ) >> 1 ;if ( a[m] >= x ) r = m ;else l = m + 1 ;}return l ;} void update ( int x , int _max , int _idx , int o , int l , int r ) {while ( l < r ) {int m = ( l + r ) >> 1 ;if ( x <= m ) {o = o << 1 ;r = m ;} else {o = o << 1 | 1 ;l = m + 1 ;}}maxv[o] = _max ;idx[o] = _idx ;while ( o > 1 ) {o >>= 1 ;if ( maxv[ls] > maxv[rs] ) {maxv[o] = maxv[ls] ;idx[o] = idx[ls] ;} else {maxv[o] = maxv[rs] ;idx[o] = idx[rs] ;}}} int query_idx , query_max ;int stack[MAXN] ; void query ( int L , int R , int o , int l , int r ) {if ( L > R ) return ;if ( L <= l && r <= R ) {if ( query_max < maxv[o] ) {query_max = maxv[o] ;query_idx = idx[o] ;}return ;}int m = mid ;if ( L <= m ) query ( L , R , lson ) ;if ( m <  R ) query ( L , R , rson ) ;} void solve () {cnt = 0 ;clr ( pre , 0 ) ;clr ( maxv , 0 ) ;clr ( idx , 0 ) ;For ( i , 1 , n ) {scanf ( "%d%d" , &node[i].a , &node[i].b ) ;node[i].idx = i ;a[++ cnt] = node[i].b ;}cnt = unique ( cnt ) ;sort ( node + 1 , node + n + 1 ) ;For ( i , 1 , n ) {int j = i ;top = 0 ;while ( j <= n && node[i].a == node[j].a ) {query_idx = 0 ;query_max = 0 ;int R = hash ( node[j].b ) ;query ( 1 , R - 1 , root ) ;pre[node[j].idx] = query_idx ;S[top ++] = Stack ( R , query_max + 1 , node[j].idx ) ;//printf ( "%d %d %d\n" , R , query_max + 1 , j ) ;++ j ;}rep ( k , 0 , top ) update ( S[k].pos , S[k].max , S[k].idx , root ) ;i = j - 1 ;}printf ( "%d\n" , maxv[1] ) ;top = 0 ;for ( int i = idx[1] ; i ; i = pre[i] ) {stack[top ++] = i ;}sort ( stack , stack + top ) ;rep ( i , 0 , top ) printf ( "%d%c" , stack[i] , i < top - 1 ? ' ' : '\n' ) ;} int main () {scanf ( "%d" , &n ) ;solve () ;return 0 ;}