文章目录

A - 2nd Greatest Distance
B - RGB Matching
C - Odd Even Sort
D - 1 or 2
E - Directed Tree
F - Logical Operations on Tree

NOMURA Programming Contest 2021(AtCoder Regular Contest 121)

A - 2nd Greatest Distance

大模拟讨论yyds

将点按x,yx,yx,y分别排序

xxx贡献最大值的点对等于yyy贡献最大值的点对

次小值就变成x/yx/yx/y中最大次小和次大最小一共四种组合中的最大值
xxx最大最小贡献点对不等于yyy最大最小点对
- xxx最大最小更大
  
  yyy最大最小与xxx中次大最小和最大次小比较
- yyy最大最小更大
  
  xxx最大最小与yyy中次大最小和最大次小比较

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define int long long
#define maxn 200005
struct node {int x, y, id;
}h1[maxn], h2[maxn];
int n;bool cmp1( node s, node t ) {return s.x < t.x;
}bool cmp2( node s, node t ) {return s.y < t.y;
}signed main() {scanf( "%lld", &n );for( int i = 1;i <= n;i ++ ) {scanf( "%lld %lld", &h1[i].x, &h1[i].y );h1[i].id = i;h2[i] = h1[i];}sort( h1 + 1, h1 + n + 1, cmp1 );sort( h2 + 1, h2 + n + 1, cmp2 );if( ( h1[1].id == h2[1].id && h1[n].id == h2[n].id ) || ( h1[1].id == h2[n].id && h1[n].id == h2[1].id ) )printf( "%lld\n", max( max( h1[n].x - h1[2].x, h1[n - 1].x - h1[1].x ), max( h2[n].y - h2[2].y, h2[n - 1].y - h2[1].y ) ) );else if( h1[n].x - h1[1].x < h2[n].y - h2[1].y )printf( "%lld\n", max( h1[n].x - h1[1].x, max( h2[n].y - h2[2].y, h2[n - 1].y - h2[1].y ) ) );elseprintf( "%lld\n", max( h2[n].y - h2[1].y, max( h1[n].x - h1[2].x, h1[n - 1].x - h1[1].x ) ) );return 0;
}

B - RGB Matching

显然，最后要么三种颜色的狗全是偶数条，两两配对，答案为000；要么只会恰有两种颜色的狗为奇数条

case 1

两种颜色中选择不满意值相差最小的两条狗

具体来说，枚举一种颜色的狗，再二分左右求出与该狗不满意值差距最小的狗，整体不满意值取最小值

case 2

第三颜色狗起一个中转点的作用

具体来说，分别枚举两种颜色的狗，以及在第三颜色中求出与枚举狗不满意值差距最小的狗

整体取不满意值最小，两种颜色最小再相加

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
#define int long long
vector < int > G[3];
int n, pos_l, pos_r;int id( char ch ) {if( ch == 'R' ) return 0;if( ch == 'B' ) return 1;if( ch == 'G' ) return 2;
}int Fabs( int x ) {return x < 0 ? -x : x;
}void work( int c, int val ) {int l = 0, r = G[c].size() - 1;pos_l = 0, pos_r = G[c].size() - 1;while( l <= r ) {int mid = ( l + r ) >> 1;if( G[c][mid] <= val ) pos_l = mid, l = mid + 1;else r = mid - 1;}l = 0, r = G[c].size() - 1;while( l <= r ) {int mid = ( l + r ) >> 1;if( G[c][mid] >= val ) pos_r = mid, r = mid - 1;else l = mid + 1;}
}signed main() {scanf( "%lld", &n );for( int i = 1;i <= ( n << 1 );i ++ ) {char c; int a;scanf( "%lld %c", &a, &c );G[id( c )].push_back( a );}int c1 = -1, c2 = -1;for( int i = 0;i < 3;i ++ )if( G[i].size() & 1 ) {if( ~ c1 ) c2 = i;else c1 = i;}if( c1 == -1 ) return ! printf( "0\n" );else {int ans = 1e18;sort( G[c2].begin(), G[c2].end() );for( int i = 0;i < G[c1].size();i ++ ) {work( c2, G[c1][i] );ans = min( ans, min( Fabs( G[c1][i] - G[c2][pos_l] ), Fabs( G[c2][pos_r] - G[c1][i] ) ) );}int ans1 = 1e18, ans2 = 1e18, c;for( int i = 0;i < 3;i ++ )if( c1 == i || c2 == i ) continue;else c = i;if( ! G[c].size() ) goto pass;sort( G[c].begin(), G[c].end() );for( int i = 0;i < G[c1].size();i ++ ) {work( c, G[c1][i] );ans1 = min( ans1, min( Fabs( G[c1][i] - G[c][pos_l] ), Fabs( G[c][pos_r] - G[c1][i] ) ) );}for( int i = 0;i < G[c2].size();i ++ ) {work( c, G[c2][i] );ans2 = min( ans2, min( Fabs( G[c2][i] - G[c][pos_l] ), Fabs( G[c][pos_r] - G[c2][i] ) ) );}pass : printf( "%lld\n", min( ans, ans1 + ans2 ) );}return 0;
}

C - Odd Even Sort

题读错做法千奇百怪错奇数次操作只能操作奇数位置，偶数次操作只能操作偶数位置

问题不要求操作最小，只求不超过n2n^2n2（冒泡排序复杂度），想法一下子就来了，明显的构造

从最大值往最小值依次考虑(kkk)，也就是只用考虑右移的操作

不难发现，如果kkk本身处于位置与操作次数恰好同奇偶，那么可以直接一直操作kkk直到位置kkk

如果不同奇偶，那么就像能否先操作一次无关的位置，再一直操作kkk

这显然是可以的，每次奇偶就让1,21,21,2位置与后面一个进行交换

当然还有位置与值已经匹配的，直接下一个哈

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 505
int T, n;
int a[maxn], b[maxn], ans[maxn * maxn];int main() {scanf( "%d", &T );while( T -- ) {scanf( "%d", &n );for( int i = 1;i <= n;i ++ )scanf( "%d", &a[i] ), b[i] = a[i];sort( b + 1, b + n + 1 );int cnt = 0, ip = n;for( int t = 1;t <= n * n && ip;t ++ ) {bool flag = 0;for( int i = 1;i < n;i ++ )if( a[i] > a[i + 1] ) {flag = 1;break;}if( ! flag ) break;int pos;for( int i = 1;i <= n;i ++ )if( a[i] == b[ip] ) {pos = i;break;}if( ( pos & 1 ) == ( t & 1 ) ) {while( pos < ip ) {ans[++ cnt] = pos;t ++;swap( a[pos], a[pos + 1] );pos ++;}t --;ip --;}else {flag = 1;for( int i = ( ( t & 1 ) ? 1 : 2 );i < n;i += 2 )if( a[i] > a[i + 1] ) {ans[++ cnt] = i;swap( a[i], a[i + 1] );flag = 0;break;}if( flag ) {int x = ( ( t & 1 ) ? 1 : 2 );ans[++ cnt] = x;swap( a[x], a[x + 1] );}}}printf( "%d\n", cnt );if( cnt ) {for( int i = 1;i <= cnt;i ++ )printf( "%d ", ans[i] );printf( "\n" );}}return 0;
}

D - 1 or 2

肯定是最大最小，次大次小…一一配对

a<b<c<d⇒max(a+d,b+c)≤max(a+c,b+d);min(a+c,b+d)≤min(a+d,b+c)a<b<c<d\Rightarrow max(a+d,b+c)\le max(a+c,b+d);min(a+c,b+d)\le min(a+d,b+c)a<b<c<d⇒max(a+d,b+c)≤max(a+c,b+d);min(a+c,b+d)≤min(a+d,b+c)

如果是一个蛋糕，不妨看做和000合并，所以只需要枚举000的个数排序顺次合并即可，O(n2)O(n^2)O(n2)

#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 5005
#define int long long
vector < int > g;
int n;signed main() {scanf( "%lld", &n );for( int i = 1, x;i <= n;i ++ ) {scanf( "%lld", &x );g.push_back( x );}sort( g.begin(), g.end() );int Size = g.size(), maxx = -1e18, minn = 1e18;for( int i = 0;i < ( Size >> 1 );i ++ ) {maxx = max( g[Size - 1 - i] + g[i], maxx );minn = min( g[Size - 1 - i] + g[i], minn );}if( Size & 1 ) {maxx = max( g[Size >> 1], maxx );minn = min( g[Size >> 1], minn );}int ans = maxx - minn;for( int i = 1;i <= n;i ++ ) {g.push_back( 0 ), Size ++;for( int i = Size - 1;i;i -- )if( g[i] < g[i - 1] ) swap( g[i], g[i - 1] );else break;maxx = -1e18, minn = 1e18;for( int i = 0;i < ( Size >> 1 );i ++ ) {maxx = max( g[Size - 1 - i] + g[i], maxx );minn = min( g[Size - 1 - i] + g[i], minn );}if( Size & 1 ) {maxx = max( g[Size >> 1], maxx );minn = min( g[Size >> 1], minn );}ans = min( ans, maxx - minn );}printf( "%lld\n", ans );return 0;
}

E - Directed Tree

套路都见过，却没有反应过来

设dpi,j:idp_{i,j}:idpi,j:i 子树内不满足条件的节点数为jjj

dpu,j+k=∑j=0sizeu∑k=0sizevdpu,j×dpv,kdp_{u,j+k}=\sum_{j=0}^{size_u}\sum_{k=0}^{size_v}dp_{u,j}\times dp_{v,k}dpu,j+k=∑j=0sizeu∑k=0sizevdpu,j×dpv,k

最后容斥

ans=∑i=0n(−1)idp1,i×(n−i)!ans=\sum_{i=0}^n(-1)^idp_{1,i}\times (n-i)!ans=∑i=0n(−1)idp1,i×(n−i)!

#include <cstdio>
#include <vector>
using namespace std;
#define int long long
#define mod 998244353
#define maxn 2005
vector < int > G[maxn];
int n;
int fac[maxn], siz[maxn], g[maxn];
int dp[maxn][maxn];void dfs( int u ) {dp[u][0] = 1;for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i];dfs( v );for( int j = 0;j <= siz[u];j ++ )for( int k = 0;k <= siz[v];k ++ )g[j + k] = ( g[j + k] + dp[u][j] * dp[v][k] % mod ) % mod;siz[u] += siz[v];for( int j = 0;j <= siz[u];j ++ )dp[u][j] = g[j], g[j] = 0;}for( int i = siz[u];~ i;i -- )dp[u][i + 1] = ( dp[u][i + 1] + dp[u][i] * ( siz[u] - i ) % mod ) % mod;siz[u] ++;
}signed main() {scanf( "%lld", &n );for( int i = 2, p;i <= n;i ++ ) {scanf( "%lld", &p );G[p].push_back( i );}dfs( 1 );fac[0] = 1;for( int i = 1;i <= n;i ++ )fac[i] = fac[i - 1] * i % mod;int ans = 0;for( int i = 0;i <= n;i ++ )if( i & 1 ) ans = ( ans - dp[1][i] * fac[n - i] % mod + mod ) % mod;else ans = ( ans + dp[1][i] * fac[n - i] % mod ) % mod;printf( "%lld\n", ans );return 0;
}

F - Logical Operations on Tree

N=1N=1N=1结果显然，考虑其他情况

如果一个叶子标记为1并且与之相连的边是OR

不管其它长什么样都可以将这条边的操作放在最后一步，从而符合要求，简称好树

接下来考虑没有这种叶子，合并出新点的情况

如果叶子u的标记为0，且相连边是AND，则这条边操作后，新点一定标记为0

如果执行这条边后的树是好树，那么原来的树也一定是好树
如果叶子u的标记为0，且相连边是OR，则这条边操作后，原来与叶子相连点的标记即为新点标记

如果执行这条边后的树是好树，那么原来的树也一定是好树
如果叶子u的标记为1，且相连边是AND，则这条边操作后，原来与叶子相连点的标记即为新点标记

如果执行这条边后的树是好树，那么原来的树也一定是好树

在树上从叶子到父亲搭建，出现1 OR时剩下的点边不管是什么，一定也是好树了；否则把叶子扔掉，父亲节点可能为0/1，在树上DPDPDP

f:f:f: 子树内没有出现1 OR想要边的树个数，$g: $ 在fff所有方案数中好树的个数

ps：没有出现1 OR这种关键叶子标记1指向父亲种类，但是可以出现1 OR父亲标记1的情况，因为可能父亲的其它儿子操作时会导致父亲标记变化

#include <cstdio>
#include <vector>
using namespace std;
#define int long long
#define mod 998244353
#define maxn 100005
vector < int > G[maxn];
int n;
int f[maxn], g[maxn];int qkpow( int x, int y ) {int ans = 1;while( y ) {if( y & 1 ) ans = ans * x % mod;x = x * x % mod;y >>= 1;}return ans;
}void dfs( int u, int fa ) {f[u] = 2, g[u] = 1;for( int i = 0;i < G[u].size();i ++ ) {int v = G[u][i];if( v == fa ) continue;else dfs( v, u );f[u] = ( ( f[v] * 2 - g[v] ) * f[u] % mod + mod ) % mod;g[u] = g[u] * f[v] % mod;}
}signed main() {scanf( "%lld", &n );for( int i = 1, u, v;i < n;i ++ ) {scanf( "%lld %lld", &u, &v );G[u].push_back( v );G[v].push_back( u );}dfs( 1, 0 );printf( "%lld\n", ( qkpow( 2, n * 2 - 1 ) - f[1] + g[1] + mod ) % mod );return 0;
}

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