【ACdream】Andrew Stankevich's Contest (22)

B：1414Geometry Problem

每两个点之间求距离。选最小的距离作为圆的直径，求出圆心和半径，最后半径加一个小小的数就能够了。

C：1415 Important Roads

对起点和终点分别求最短路，得到一定在最短路上的边（满足d[s][u]+w[u][v]+d[u][t]==d[s][t]的边），然后用双连通求桥的方法求边就可以。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
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 )
 
const int MAXN = 20005 ;
const int MAXE = 400005 ;
const LL INF = 1e18 ;
 
struct Edge {
	int v , c ;
	Edge* next ;
} E[MAXE] , *H[MAXN] , *edge ;
 
struct Seg {
	int u , v , c ;
	Seg () {}
	Seg ( int u , int v , int c ) : u ( u ) , v ( v ) , c ( c ) {}
} seg[MAXE] ;
 
struct Node {
	LL d ;
	int idx ;
	Node () {}
	Node ( LL d , int idx ) : d ( d ) , idx ( idx ) {}
	bool operator < ( const Node& a ) const {
		return d > a.d ;
	}
} ;
 
LL dis[2][MAXN] ;
int vis[MAXN] , Time ;
int Q[MAXN] , head , tail ;
int n ;
 
void clear () {
	edge = E ;
	clr ( H , 0 ) ;
}
 
void addedge ( int u , int v , int c ) {
	edge->v = v ;
	edge->c = c ;
	edge->next = H[u] ;
	H[u] = edge ++ ;
}
 
void dijkstra ( int s , LL d[] ) {
	For ( i , 1 , n ) d[i] = INF ;
	priority_queue < Node > q ;
	++ Time ;
	d[s] = 0 ;
	q.push ( Node ( d[s] , s ) ) ;
	while ( !q.empty () ) {
		int u = q.top ().idx ;
		q.pop () ;
		if ( vis[u] == Time ) continue ;
		vis[u] = Time ;
		travel ( e , H , u ) {
			int v = e->v ;
			if ( d[v] > d[u] + e->c ) {
				d[v] = d[u] + e->c ;
				q.push ( Node ( d[v] , v ) ) ;
			}
		}
	}
}
 
struct BCC {
	Edge E[MAXE] , *H[MAXN] , *edge ;
	int dfn[MAXN] , low[MAXN] ;
	int dfs_clock ;
	int ans[MAXN] , top ;
	 
	void clear () {
		edge = E ;
		dfs_clock = 0 ;
		clr ( H , 0 ) ;
		clr ( dfn , 0 ) ;
		clr ( low , 0 ) ;
	}
	 
	void addedge ( int u , int v , int c ) {
		edge->v = v ;
		edge->c = c ;
		edge->next = H[u] ;
		H[u] = edge++ ;
	}
	 
	void tarjan ( int u , int fa = 0 ) {
		dfn[u] = low[u] = ++ dfs_clock ;
		int flag = 1 ;
		travel ( e , H , u ) {
			int v = e->v ;
			if ( v == fa && flag ) {
				flag = 0 ;
				continue ;
			}
			if ( !dfn[v] ) {
				tarjan ( v , u ) ;
				low[u] = min ( low[u] , low[v] ) ;
				if ( low[v] > dfn[u] ) ans[top ++] = e->c ;
			} else low[u] = min ( low[u] , dfn[v] ) ;
		}
	}
	 
	void find_bcc ( int n ) {
		top = 0 ;
		For ( i , 1 , n ) if ( !dfn[i] ) tarjan ( i ) ;
		printf ( "%d
" , top ) ;
		sort ( ans , ans + top ) ;
		rep ( i , 0 , top ) printf ( "%d%c" , ans[i] , i < top - 1 ? ' ' : '
' ) ;
	}
} G ;
 
int m ;
 
void scanf ( int& x , char c = 0 ) {
	while ( ( c = getchar () ) < '0' || c > '9' ) ;
	x = c - '0' ;
	while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ;
}
 
void solve () {
	int u , v , c ;
	clear () ;
	G.clear () ;
	For ( i , 1 , m ) {
		scanf ( u ) , scanf ( v ) , scanf ( c ) ;
		addedge ( u , v , c ) ;
		addedge ( v , u , c ) ;
		seg[i] = Seg ( u , v , c ) ;
	}
	dijkstra ( 1 , dis[0] ) ;
	dijkstra ( n , dis[1] ) ;
	LL tot_dis = dis[0][n] ;
	For ( i , 1 , m ) {
		u = seg[i].u ;
		v = seg[i].v ;
		if ( dis[0][u] + seg[i].c + dis[1][v] == tot_dis || dis[0][v] + seg[i].c + dis[1][u] == tot_dis ) {
			G.addedge ( u , v , i ) ;
			G.addedge ( v , u , i ) ;
		}
	}
	G.find_bcc ( n ) ;
}

int main () {
	clr ( vis , 0 ) ;
	Time = 0 ;
	while ( ~scanf ( "%d%d" , &n , &m ) ) solve () ;
	return 0 ;
}

E：1417 Numbers

对该数求全部比他大的小于等于n的开头为1的最小的数（这个数满足：10^i +（k -（10^i - 1）% k + 1））。

#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 )
 
const int MAXN = 65 ;
const int MAXE = 10005 ;
const int MAXH = 10005 ;
 
LL n , k ;
LL a[MAXN] ;
 
struct String {
	char s[MAXN] ;
} s[MAXN] ;
int cnt = 0 ;
 
LL pow ( int a , int b ) {
	LL res = 1 , tmp = a ;
	while ( b ) {
		if ( b & 1 ) res *= tmp ;
		tmp *= tmp ;
		b >>= 1 ;
	}
	return res ;
}
 
int cmp ( const String& a , const String& b ) {
	return strcmp ( a.s , b.s ) < 0 ;
}
 
void solve () {
	cnt = 0 ;
	a[cnt ++] = k ;
	For ( i , 1 , 18 ) {
		LL tmp = pow ( 10 , i ) ;
		if ( tmp > n ) break ;
		tmp = tmp + ( k - ( tmp - 1 ) % k - 1 ) ;
		if ( tmp <= k ) continue ;
		if ( tmp > n ) break ;
		a[cnt ++] = tmp ;
	}
	rep ( i , 0 , cnt ) sprintf ( s[i].s , "%lld" , a[i] ) ;
	sort ( s , s + cnt , cmp ) ;
	printf ( "%s
" , s[0].s ) ;
}
 
int main () {
	while ( ~scanf ( "%lld%lld" , &n , &k ) && ( n || k ) ) solve () ;
	return 0 ;
}

Ｈ：1420 High Speed Trains

设f（n）为n个城市时满足条件的方案数。

则f（n）= 全部方案数 - 有一个点没有边覆盖的方案数 - 有两个点没有边覆盖的方案数 - …… - 全都点都没有被覆盖的方案数 = 2^(n*(n-1)/2) - C[n][1] * f(n-1) - C[n][2] * f(n-2) - …… - C[n][n] * f(0)。

C[i][j]表示从i个里面选j个的组合数。

边界条件为f(1)=0，f(0)=1。

JAVA打表AC。

I：问怎样选择退出的时刻和买保险的时刻来使得期望获得的钱最多。

枚举结束的时间。每一个时间里面枚举买保险的时间（能够不买）。最后随便搞搞即可了。

有O（N）的做法。只是懒得搞了。懂了即可了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
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 )
  
const int MAXN = 55 ;
const double eps = 1e-8 ;
  
double dp[MAXN][MAXN] ;
double p[MAXN] ;
double pp[MAXN] ;
LL two[MAXN] ;
int n , c ;
  
int dcmp ( double x ) {
	return ( x > eps ) - ( x < -eps ) ;
}
  
void solve () {
	pp[0] = 1 ;
	two[0] = 1 ;
	double ans = 100 ;
	clr ( dp , 0 ) ;
	For ( i , 1 , n ) scanf ( "%lf" , &p[i] ) ;
	For ( i , 1 , n ) p[i] /= 100 ;
	For ( i , 1 , n ) pp[i] = pp[i - 1] * p[i] ;
	For ( i , 1 , n ) two[i] = two[i - 1] * 2 ;
	For ( i , 1 , n ) {
		For ( j , 1 , i ) {
			if ( 100 * two[j - 1] <= c ) continue ;
			double tmp = pp[j - 1] ;
			double tmp_p = 1 ;
			For ( k , j + 1 , i ) tmp_p *= p[k] * 2 ;
			tmp *= ( 100 * two[j - 1] - c ) * ( ( 1 - p[j] ) * tmp_p + p[j] * 2 * tmp_p ) ;
			ans = max ( ans , tmp ) ;
		}
		ans = max ( ans , 100 * two[i] * pp[i] ) ;
	}
	printf ( "%.10f
" , ans ) ;
}
  
int main () {
	while ( ~scanf ( "%d%d" , &n , &c ) ) solve () ;
	return 0 ;
}