Prob. 1
Desc. & Link.
行走的形式是比较自由的，因为只要走到了最优答案处就可以不管了，所以不需要考虑游戏的结束。
考虑二分答案。
然后预处理出每个节点到 $s$（另一棵树就是 $t$）的距离。
判断答案合法性：
首先枚举 $A$ 树的每个节点（节点编号小于当前二分的答案），然后判断需要构成答案的 $B$ 树上的节点距离 $t$ 的距离的奇偶性是否一致即可。
但是这样有一个问题：我们如何确保这个答案是整个一轮移动过程中最大的？
仔细考虑一下，我们判断成功的依据是行走过程中所有和不超过我们当前二分的值，那么转为判断这个问题（意思就是前面降智了）。
因为这是一棵树，所以该走的路径一定会走。
因为我们枚举了 $A$ 树中的节点，所以每次从哪两个节点走到 $s$、$t$ 是固定下来的。
草，直接 bfs 判断找可达最小值就行了。
$\Theta(n\log_{2}^{2}n)$，我觉得不行，先写。
草卡卡常过了。
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include <cstdio>
#include <queue>
using namespace std;
const int MAXN = 1e6 + 5;
namespace IO{
const int sz=1<<22;
char a[sz+5],b[sz+5],*p1=a,*p2=a,*t=b,p[105];
inline char gc(){
return p1==p2?(p2=(p1=a)+fread(a,1,sz,stdin),p1==p2?EOF:*p1++):*p1++;
}
template<class T> void gi(T& x){
x=0; char c=gc();
for(;c<'0'||c>'9';c=gc());
for(;c>='0'&&c<='9';c=gc())
x=x*10+(c-'0');
}
inline void flush(){fwrite(b,1,t-b,stdout),t=b; }
inline void pc(char x){*t++=x; if(t-b==sz) flush(); }
template<class T> void pi(T x,char c='\n'){
if(x==0) pc('0'); int t=0;
for(;x;x/=10) p[++t]=x%10+'0';
for(;t;--t) pc(p[t]); pc(c);
}
struct F{~F(){flush();}}f;
}
using IO::gi;
using IO::pi;
template<typename _T> _T MIN ( const _T x, const _T y ) { return x < y ? x : y; }
struct GraphSet {
int to, nx;
GraphSet ( int T = 0, int N = 0 ) { to = T, nx = N; }
} asA[MAXN * 2], asB[MAXN * 2];
int n, s, t, cntA, cntB, beginA[MAXN], beginB[MAXN], disA[MAXN], disB[MAXN], visA[MAXN], visB[MAXN];
void makeEdgeA ( const int u, const int v ) { asA[++ cntA] = GraphSet ( v, beginA[u] ), beginA[u] = cntA; }
void makeEdgeB ( const int u, const int v ) { asB[++ cntB] = GraphSet ( v, beginB[u] ), beginB[u] = cntB; }
void dfsA ( const int u, const int lst, const int dis ) {
disA[u] = dis;
for ( int i = beginA[u]; i; i = asA[i].nx ) {
int v = asA[i].to;
if ( v == lst ) continue;
dfsA ( v, u, dis + 1 );
}
}
void dfsB ( const int u, const int lst, const int dis ) {
disB[u] = dis;
for ( int i = beginB[u]; i; i = asB[i].nx ) {
int v = asB[i].to;
if ( v == lst ) continue;
dfsB ( v, u, dis + 1 );
}
}
void behaveOneSide ( int ark, int& mnA, int& mnB, int& ord, priority_queue<int, vector<int>, greater<int>>& align ) {
int preSave = mnA;
while ( ! align.empty () ) {
int u = align.top ();
if ( u + mnB > ark ) break;
else align.pop ();
for ( int i = beginA[u]; i; i = asA[i].nx ) {
int v = asA[i].to;
if ( visA[v] ) continue;
visA[v] = 1, align.push ( v );
mnA = MIN ( mnA, v );
}
}
if ( mnA == preSave ) ++ ord;
else ord = 0;
}
void behaveAnotherSide ( int ark, int& mnA, int& mnB, int& ord, priority_queue<int, vector<int>, greater<int>>& align ) {
int preSave = mnB;
while ( ! align.empty () ) {
int u = align.top ();
if ( u + mnA > ark ) break;
else align.pop ();
for ( int i = beginB[u]; i; i = asB[i].nx ) {
int v = asB[i].to;
if ( visB[v] ) continue;
visB[v] = 1, align.push ( v );
mnB = MIN ( mnB, v );
}
}
if ( mnB == preSave ) ++ ord;
else ord = 0;
}
priority_queue<int, vector<int>, greater<int>> oneQ, anotherQ;
bool check ( const int x ) {
for ( int i = 1; i <= n; ++ i ) visA[i] = visB[i] = 0;
for ( ; ! oneQ.empty (); oneQ.pop () ) ;
for ( ; ! anotherQ.empty (); anotherQ.pop () ) ;
oneQ.push ( s ), anotherQ.push ( t );
visA[s] = 1, visB[t] = 1;
int turn = 0, mnA = s, mnB = t, ord = 0;
while ( mnA > 1 || mnB > 1 ) {
turn ^= 1;
if ( turn ) behaveOneSide ( x, mnA, mnB, ord, oneQ );
else behaveAnotherSide ( x, mnA, mnB, ord, anotherQ );
if ( ord > 2 ) break;
}
if ( mnA == 1 && mnB == 1 ) return 1;
else return 0;
}
int solve ( int l, int r ) {
while ( l + 1 < r ) {
int mid = ( l + r ) >> 1;
if ( check ( mid ) ) r = mid;
else l = mid;
}
return r;
}
int main () {
int tCase;
gi ( tCase );
while ( tCase -- > 0 ) {
gi ( n ), cntA = cntB = 0;
for ( int i = 1; i <= n; ++ i ) beginA[i] = 0, beginB[i] = 0;
for ( int i = 1, u, v; i < n; ++ i ) {
gi ( u ), gi ( v );
makeEdgeA ( u, v ), makeEdgeA ( v, u );
}
for ( int i = 1, u, v; i < n; ++ i ) {
gi ( u ), gi ( v );
makeEdgeB ( u, v ), makeEdgeB ( v, u );
}
gi ( s ), gi ( t );
// dfsA ( s, 0, 0 ), dfsB ( t, 0, 0 );
pi ( solve ( 1, n << 1 ) );
}
return 0;
}
Prob. 2
Desc. & Link.
$n$ 很小，$q$ 也很小。
感觉这个 $n$ 不是 $2^{n}$ 的算法也不是多项式算法欸。
但复杂度一定与 $n$ 有关……
草这玩意儿折半是不是可以折半搜索？
我们可以搜出两边我们可以凑出的价格，分别记为 $A_{i},i\in[1,C_{A}]$、$B_{i},i\in[1,C_{B}]$。
然后让 $A,B$ sorted。
然后枚举 $A_{i}$，找到 $B$ 中最大的能与 $A_{i}$ 相加小于等于需要的值，然后算下贡献即可（bullshit）。
不是为什么用快读本地过数据提交瓦爆啊。。。
#include<cstdio>
#include<algorithm>
using namespace std;
void read(long long &hhh)
{
long long x=0;
char c=getchar();
while(((c<'0')|(c>'9'))&(c^'-')) c=getchar();
if(c^'-') x=c^'0';
char d=getchar();
while((d>='0')&(d<='9'))
{
x=(x<<3)+(x<<1)+(d^'0');
d=getchar();
}
if(c^'-') hhh=x;
else hhh=-x;
}
void writing(long long x)
{
if(!x) return;
if(x>9) writing(x/10);
putchar((x%10)^'0');
}
void write(long long x)
{
if(x<0)
{
putchar('-');
x=-x;
}
else if(!x)
{
putchar('0');
putchar('\n');
return;
}
writing(x);
putchar('\n');
}
long long n,q,endone,beginano,onesiz,onebuc[2000005],anosiz,anobuc[2000005],opl,opr,cud[45];
void dfs(long long now,long long cur)
{
if(now==endone+1) onebuc[++onesiz]=cur;
else
{
dfs(now+1,cur+cud[now]);
dfs(now+1,cur);
}
}
void exdfs(long long now,long long cur)
{
if(now==n+1) anobuc[++anosiz]=cur;
else
{
exdfs(now+1,cur+cud[now]);
exdfs(now+1,cur);
}
}
long long solve(long long mos)
{
long long now=anosiz;
long long res=0;
for(long long i=1;i<=onesiz;++i)
{
while(now&&onebuc[i]+anobuc[now]>mos) now--;
res+=now;
}
return res;
}
int main()
{
// read(n);
// read(q);
scanf("%lld%lld",&n,&q);
// for(long long i=1;i<=n;++i) read(cud[i]);
for(long long i=1;i<=n;++i) scanf("%lld",&cud[i]);
endone=(n>>1);
beginano=endone+1;
dfs(1,0);
exdfs(beginano,0);
sort(onebuc+1,onebuc+onesiz+1);
sort(anobuc+1,anobuc+anosiz+1);
while(q--)
{
scanf("%lld%lld",&opl,&opr);
// read(opl);
// read(opr);
// write(solve(opr)-solve(opl-1));
printf("%lld\n",solve(opr)-solve(opl-1));
}
return 0;
}
Prob. 4
Desc. & Link.
相当于在树上对于每一个点找出找出一条以其为链顶的链。
设 $f_{u}$ 为 $u$ 的答案。
$$
f_{u}=\max_{v\in\text{son}(u)}\{f_{v}+(a_{u}-\text{dis}(u,v))\times b_{v},0\}
$$
有乘法，然后题目中一堆常数。
:-) 斜率优化
我们令 $s_{u}=\text{dis}(1,u)$，然后
$$
\begin{aligned}
f_{u}
&=\max_{v\in\text{son}(u)}\{f_{v}+(a_{u}+s_{u}-s_{v})\times b_{v},0\} \\
&=\max_{v\in\text{son}(u)}\{(a_{u}-s_{u})\times b_{v}+f_{v}-s_{v}\times b_{v},0\}
\end{aligned}
$$
令 $y=f_{u},x=a_{u}-s_{u},k=b_{v},b=f_{v}-s_{v}\times b_{v}$，那么这个东西就是一个 $y=kx+b$。
那么我们现在需要在子树里维护凸包，并且能够支持合并凸包和插入直线。