#699-Div.2

A. Space Navigation

对于最终位置(x,y)，我们只关心那两个方向的字符是否足够即可

#include <cstdio>
#include <cstring>
#define maxn 100005
int T, n, m;
char s[maxn];int main() {scanf( "%d", &T );while( T -- ) {scanf( "%d %d %s", &n, &m, s + 1 );char x = n < 0 ? 'L' : 'R';char y = m < 0 ? 'D' : 'U';int len = strlen( s + 1 );int cntx = 0, cnty = 0;for( int i = 1;i <= len;i ++ ) {if( s[i] == x ) cntx ++;if( s[i] == y ) cnty ++;}if( n < 0 ) n = -n;if( m < 0 ) m = -m;if( cntx >= n && cnty >= m ) printf( "YES\n" );else printf( "NO\n" );}return 0;
}

B. New Colony

发现数据非常小，最暴力也只用滚$1e5$次就一定是$-1$，所以$k$那么大纯粹虾银

#include <cstdio>
#define maxn 105
int T, n, k;
int h[maxn];int main() {scanf( "%d", &T );while( T -- ) {scanf( "%d %d", &n, &k );for( int i = 1;i <= n;i ++ )scanf( "%d", &h[i] );int ans;for( int i = 1;i <= k;i ++ ) {int pos = 1;for( int j = 2;j <= n;j ++ )if( h[pos] >= h[j] ) pos ++;else {h[pos] ++;break;}ans = pos;if( ans == n ) break;}if( ans == n ) printf( "-1\n" );else printf( "%d\n", ans );}	return 0;
}

C. Fence Painting

统计需要染色的颜色和个数，判断是否足够

对于无效的笔又必须使用，显然将之用到最后一次有效操作的木板上

这样之前的所有染色都会被覆盖掉

注意最后一次有效操作的模板查找

#include <cstdio>
#include <queue>
using namespace std;
#define maxn 100005
queue < int > q[maxn];
int T, n, m;
int a[maxn], b[maxn], c[maxn], cnt[maxn];
bool vis[maxn];int main() {scanf( "%d", &T );again :while( T -- ) {scanf( "%d %d", &n, &m );for( int i = 1;i <= n;i ++ ) {while( ! q[i].empty() ) q[i].pop();vis[i] = cnt[i] = 0;}for( int i = 1;i <= n;i ++ )scanf( "%d", &a[i] );for( int i = 1;i <= n;i ++ )scanf( "%d", &b[i] ), vis[b[i]] = 1;for( int i = 1;i <= m;i ++ )scanf( "%d", &c[i] ), cnt[c[i]] ++;if( ! vis[c[m]] ) {printf( "NO\n" );goto again;}int pos;for( int i = 1;i <= n;i ++ ) {if( a[i] != b[i] ) q[b[i]].push( i );if( b[i] == c[m] && ( a[i] != b[i] || ( a[i] == b[i] && ! q[b[i]].size() ) ) ) pos = i;}for( int i = 1;i <= n;i ++ )if( q[b[i]].size() > cnt[b[i]] ) {printf( "NO\n" );goto again;}printf( "YES\n" );for( int i = 1;i <= m;i ++ )if( ! q[c[i]].empty() ) {printf( "%d ", q[c[i]].front() );q[c[i]].pop();}else printf( "%d ", pos );printf( "\n" );}return 0;
}

D. AB Graph

大讨论就是猴赛雷！！！

• $m=2k+1$

  aba aaa都是回文串，由此可知不管边长什么样随便选两个点都是正确的

  在这里我们采取在(1,2)之间反复横跳

• $m=2k$

  • 先特判一下$n=2$的情况

  • 存在$E_{a\to b}=E_{b\to a}$，那么只需要在这样一对边横跳即可，回文串只包含了一个字符

  • 一个简单的三元环所经过的路径字符一样，直接绕环走

  • 不存在那么完美的两种情况，这意味着一对边一定是ab类型的

    显然，任意三个点中一定会有至少一个点连出的两条边为ab

    （当控制另外两个点连出边一样时，第三个点的边就已确定）

    • $\frac{m}{2}=3t$，走法bab
    • $\frac{m}{2}=3t+1$，走法abb
    • $\frac{m}{2}=3t+2$，走法bba

PS：代码虽然看着冗长，但本质是ctrl+c ctrl+v (#^.#)

#include <cstdio>
#include <iostream>
using namespace std;
#define maxn 1005
int T, n, m;
bool flag;
char s[maxn][maxn];bool check0( int i, int j, int k ) {if( s[i][j] == 'b' && s[j][k] == 'a' && s[k][i] == 'b' ) return 1;else return 0;
}bool check1( int i, int j, int k ) {if( s[i][j] == 'b' && s[j][k] == 'b' && s[k][i] == 'a' ) return 1;else return 0;
}bool check2( int i, int j, int k ) {if( s[i][j] == 'a' && s[j][k] == 'b' && s[k][i] == 'b' ) return 1;else return 0;
}void print( int i, int j, int k ) {printf( "YES\n" );for( int p = 0;p <= ( m << 1 );p ++ ) {printf( "%d ", i );int t = i;i = j;j = k;k = t;}printf( "\n" );flag = 1;
}bool Check( int i, int j, int k ) {if( s[i][j] == s[j][k] && s[j][k] == s[k][i] ) return 1;else return 0;
}int check( int i, int j ) {for( int k = 1;k <= n;k ++ )if( s[i][j] == s[j][k] ) return k;return 0;
}void Print( int i, int j, int k ) {flag = 1;printf( "YES\n%d %d %d\n", i, j, k );
}int main() {scanf( "%d", &T );again :while( T -- ) {scanf( "%d %d", &n, &m );for( int i = 1;i <= n;i ++ )scanf( "%s", s[i] + 1 );if( m & 1 ) {int now = 1;printf( "YES\n" );for( int i = 0;i <= m;i ++ ) {printf( "%d ", now );now = 3 - now;}printf( "\n" );goto again;}else {for( int i = 1;i < n;i ++ )for( int j = i + 1;j <= n;j ++ )if( s[i][j] == s[j][i] ) {printf( "YES\n" );for( int k = 0;k <= m;k ++ ) {printf( "%d ", i );swap( i, j );}printf( "\n" );goto again;}if( n == 2 ) {printf( "NO\n" );goto again;}m >>= 1;flag = 0;if( m == 1 ) {int pos;if( pos = check( 1, 2 ) ) Print( 1, 2, pos );else if( pos = check( 1, 3 ) ) Print( 1, 3, pos );else if( pos = check( 2, 1 ) ) Print( 2, 1, pos );else if( pos = check( 2, 3 ) ) Print( 2, 3, pos );else if( pos = check( 3, 1 ) ) Print( 3, 1, pos );else if( pos = check( 3, 2 ) ) Print( 3, 2, pos );}else {flag = 0;int k = m % 3;switch( k ) {case 0 : {if( Check( 1, 2, 3 ) ) print( 1, 2, 3 );else if( check0( 1, 2, 3 ) ) print( 1, 2, 3 );else if( check0( 1, 3, 2 ) ) print( 1, 3, 2 );else if( check0( 2, 1, 3 ) ) print( 2, 1, 3 );else if( check0( 2, 3, 1 ) ) print( 2, 3, 1 );else if( check0( 3, 1, 2 ) ) print( 3, 1, 2 );else if( check0( 3, 2, 1 ) ) print( 3, 2, 1 );break;}case 1 : {if( Check( 1, 2, 3 ) ) print( 1, 2, 3 );else if( check1( 1, 2, 3 ) ) print( 1, 2, 3 );else if( check1( 1, 3, 2 ) ) print( 1, 3, 2 );else if( check1( 2, 1, 3 ) ) print( 2, 1, 3 );else if( check1( 2, 3, 1 ) ) print( 2, 3, 1 );else if( check1( 3, 1, 2 ) ) print( 3, 1, 2 );else if( check1( 3, 2, 1 ) ) print( 3, 2, 1 );break;}case 2 : {if( Check( 1, 2, 3 ) ) print( 1, 2, 3 );else if( check2( 1, 2, 3 ) ) print( 1, 2, 3 );else if( check2( 1, 3, 2 ) ) print( 1, 3, 2 );else if( check2( 2, 1, 3 ) ) print( 2, 1, 3 );else if( check2( 2, 3, 1 ) ) print( 2, 3, 1 );else if( check2( 3, 1, 2 ) ) print( 3, 1, 2 );else if( check2( 3, 2, 1 ) ) print( 3, 2, 1 );break;}}}if( ! flag ) printf( "NO\n" );}}return 0;
}

E. Sorting Books

如果直接去求操作谁，显然要考虑操作数量以及顺序问题，顺序问题一般都非常吃力不讨好

反过来从后往前考虑，去求不操的数量的最大值即可，这时候就不需要管顺序了

设$d{p}_i:[i,n]$ 最大不操次数，$l_i,r_i:i$ 值出现的最左点和最右点，$cn{t}_c:$ 到$i$位置为止$c$值出现的个数

• 现在位置$i$上的值的最左点就在$i$处，那么贪心的可以选择$[i,n]$所有与$a_i$相同点不操作

  $d{p}_{r_{a_i}+1}+cn{t}_{a_i}$

• 最左点不在$i$处，贪心的也可以选择$[i,n]$所有出现的与$a_i$相同点不操作

  $cn{t}_{a_i}$

• 位置$i$操作

  $d{p}_{i+1}$

#include <cstdio>
#include <iostream>
using namespace std; 
#define maxn 500005
int n;
int a[maxn], l[maxn], r[maxn], cnt[maxn], dp[maxn];int main() {scanf( "%d", &n );for( int i = 1;i <= n;i ++ ) {scanf( "%d", &a[i] );if( ! l[a[i]] ) l[a[i]] = i;r[a[i]] = i;}for( int i = n;i;i -- ) {++ cnt[a[i]];dp[i] = dp[i + 1];if( i == l[a[i]] )dp[i] = max( dp[i], dp[r[a[i]] + 1] + cnt[a[i]] );elsedp[i] = max( dp[i], cnt[a[i]] );}printf( "%d\n", n - dp[1] );return 0;
}

F. AB Tree

• 每层点赋一样的，可以使答案最优

  可以把每一层看成物品，物品的价值就是这一层的点数，跑一个01背包

  如果存在一组物品使得它们的和恰好为$x$，那么就直接输出层数即可

• 非要修改的话，修改单层的叶子结点，可以使答案最优

  先找到叶子节点最多的一层，试图通过改当前层的叶子节点的状态来弥补$DP$状态

在此基础上，需要套用bitset优化即可卡过去

#include <cstdio>
#include <vector>
#include <bitset>
#include <iostream>
using namespace std;
#define maxn 100005
#define maxk 3000
vector < int > G[maxn], D[maxn];
int n, x, m;
int dep[maxn], num[maxn], id[maxn], w[maxn], leaf_cnt[maxn];
bool vis[maxn];
bitset < maxn > f[maxk];void dfs1( int u, int fa ) {dep[u] = dep[fa] + 1;m = max( m, dep[u] );num[dep[u]] ++;if( G[u].size() == 1 ) leaf_cnt[dep[u]] ++;for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i];if( v == fa ) continue;else dfs1( v, u );}
}void dfs2( int x, int y ) {if( ! x ) return;for( int i = 0;i < D[x].size();i ++ ) {if( w[x] > y || f[x - 1][y] ) break;y -= w[x];vis[D[x][i]] = 1;}dfs2( x - 1, y );
}int main() {scanf( "%d %d", &n, &x );for( int i = 2, fa;i <= n;i ++ ) {scanf( "%d", &fa );G[fa].push_back( i );G[i].push_back( fa );}dfs1( 1, 0 );int tot = 0;for( int i = 1;i <= n;i ++ ) {//merge depth i who has the same nodes in numbersif( ! id[num[i]] ) {id[num[i]] = ++ tot;w[tot] = num[i];D[tot].push_back( i );}else D[id[num[i]]].push_back( i );}f[0][0] = 1;for( int i = 1;i <= tot;i ++ ) {f[i] = f[i - 1];int siz = D[i].size();for( int j = 1;j <= siz;j <<= 1 ) {siz -= j;f[i] |= f[i] << j * w[i];}if( siz ) f[i] |= f[i] << siz * w[i];}if( f[tot][x] ) {printf( "%d\n", m );dfs2( tot, x );for( int i = 1;i <= n;i ++ )if( vis[dep[i]] ) printf( "a" );else printf( "b" );}else {printf( "%d\n", m + 1 );int ans;for( int i = x;~ i;i -- )if( f[tot][i] ) {ans = i;break;}dfs2( tot, ans );int pos = -1;for( int i = 1;i <= m;i ++ )if( ! vis[i] && leaf_cnt[i] >= x - ans ) {pos = i;break;}for( int i = 1;i <= n;i ++ )if( dep[i] == pos && G[i].size() == 1 ) {if( ans == x ) printf( "b" );else printf( "a" ), ans ++;}else {if( vis[dep[i]] ) printf( "a" );else printf( "b" );}}return 0;
}