2023冲刺国赛模拟6
A.A
缩成 \(ABABA..\) 每次删去两个，于是猜结论，取前 \((n - 1) / 2\) 大
code
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
int read(){
int x = 0; char c = getchar();
while(!isdigit(c))c = getchar();
do{x = x * 10 + (c ^ 48); c = getchar();}while(isdigit(c));
return x;
}
const int maxn = 1e6 + 55;
int n, w[maxn], top;
char s[maxn];
int main(){
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
n = read(); scanf("%s",s + 1);
for(int i = 1; i <= n; ++i)w[i] = read();
top = 1;
for(int i = 2; i <= n; ++i)if(s[i] == s[i - 1])w[top] = max(w[top], w[i]); else w[++top] = w[i];
if(top <= 2){printf("0\n");return 0;}
int cnt = (top - 1) / 2;
sort(w + 2, w + top);
ll ans = 0;
for(int i = 1; i <= cnt; ++i)ans += w[top - i];
printf("%lld\n",ans);
return 0;
}
B.B
对操作分块，处理 \(f_{i, j}\) 表示经过第 \(i\) 个块后 \(j\) 会到哪个位置
散块暴力搞
考虑如何处理一个块的信息
把涉及到的点拿出来建立虚树，两个点之间路径上的点用双端队列维护
对其他点处理在 \(t\) 时刻会出现在虚树上位置 \(x\)
用并查集（其实是个链表） 把重合的点缩一块，简单讨论每个位置是向上走还是向下走即可
写起来比较愉悦身心
不长，真的不长
code
#include<bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
int read(){
int x = 0; char c = getchar();
while(!isdigit(c))c = getchar();
do{x = x * 10 + (c ^ 48); c = getchar();}while(isdigit(c));
return x;
}
const int maxn = 1e5 + 5, sq = 450;
int n, m, q, fa[maxn], a[maxn], B;
struct graph{
int head[maxn], e[maxn];
void add(int u, int v){
e[v] = head[u];
head[u] = v;
}
}t, it;
int son[maxn], si[maxn], dep[maxn], top[maxn], dfn[maxn], id[maxn], tim;
void dfs1(int x){
si[x] = 1;
for(int v = t.head[x]; v; v = t.e[v]){
dep[v] = dep[x] + 1;
dfs1(v); si[x] += si[v];
if(si[son[x]] < si[v])son[x] = v;
}
}
void dfs2(int x, int tp){
dfn[x] = ++tim; id[tim] = x; top[x] = tp;
if(son[x])dfs2(son[x], tp);
for(int v = t.head[x]; v; v = t.e[v])
if(v != son[x])dfs2(v, v);
si[x] = si[x] + dfn[x] - 1;
}
int LCA(int u, int v){
while(top[u] != top[v]){
if(dep[top[u]] > dep[top[v]])u = fa[top[u]];
else v = fa[top[v]];
}
return dep[u] < dep[v] ? u : v;
}
int kfa(int x, int k){
int tmp = dfn[x] - dfn[top[x]] + 1;
while(k >= tmp){
k -= tmp; x = fa[top[x]];
tmp = dfn[x] - dfn[top[x]] + 1;
}
return id[dfn[x] - k];
}
int move(int x, int y){
if(x == y)return x;
if(dfn[x] <= dfn[y] && dfn[y] <= si[x])return kfa(y, dep[y] - dep[x] - 1);
return fa[x];
}
int tr[sq][maxn], tres[maxn], L[sq], R[sq], bl[maxn];
deque<int>que[maxn];
int node[maxn + maxn], tot, thq[maxn];
bool cmp(int x, int y){return dfn[x] < dfn[y];}
struct dsu{
int f[maxn];
void init(){for(int i = 1; i <= n; ++i)f[i] = i;}
int fa(int x){return f[x] == x ? x : f[x] = fa(f[x]);}
void merge(int x, int y){f[fa(y)] = x;}
}S;
vector<pii>rmg[sq];
int mxlen, nfa[maxn], nid[maxn];
void dfs(int x, int len, int las){
if(len == mxlen)return;
if(!thq[x])rmg[len].emplace_back(x, las);
for(int v = t.head[x]; v; v = t.e[v]){
if(thq[v])dfs(v, 0, v);
else dfs(v, len + 1, las);
}
}
void move_to(int x, int &pos){
if(x == 0)return;
if(pos == 0)pos = x;
else S.merge(pos, x);
}
void change(int x, int ban){
for(int v = it.head[x]; v; v = it.e[v])if(v != ban){
if(que[v].size()){
move_to(que[v].back(), nid[x]);
que[v].pop_back();
que[v].push_front(nid[v]);
}else move_to(nid[v], nid[x]);
nid[v] = 0; change(v, 0);
}
}
void moveto(int v){
int x = 0;
while(true){
change(v, x);
x = v; v = nfa[v];
if(v){
if(que[x].size()){
move_to(que[x].front(), nid[x]);
que[x].pop_front();
que[x].push_back(nid[v]);
}else move_to(nid[v], nid[x]);
nid[v] = 0;
}else break;
}
}
void get_tr(int bl){
tot = 0; mxlen = R[bl] - L[bl] + 1;
for(int i = L[bl]; i <= R[bl]; ++i)node[++tot] = a[i];
sort(node + 1, node + tot + 1, cmp);
for(int i = tot; i > 1; --i)node[++tot] = LCA(node[i], node[i - 1]);
node[++tot] = 1; sort(node + 1, node + tot + 1, cmp);
tot = unique(node + 1, node + tot + 1) - node - 1;
for(int i = 1; i <= tot; ++i)nid[node[i]] = node[i], thq[node[i]] = node[i];
for(int i = 2; i <= tot; ++i){
int x = fa[node[i]];
while(!thq[x])que[node[i]].push_back(x), thq[x] = node[i], x = fa[x];
nfa[node[i]] = x;
it.add(x, node[i]);
}
dfs(1, 0, 1); S.init();
for(int i = L[bl]; i <= R[bl]; ++i){
int rt = a[i]; moveto(rt);
for(pii v : rmg[i - L[bl] + 1]){
int iq = thq[v.second];
if(iq == v.second)move_to(v.first, nid[iq]);
else move_to(v.first, que[iq][dep[iq] - dep[v.second] - 1]);
}
rmg[i - L[bl] + 1].clear();
}
for(int i = 2; i <= tot; ++i){
int x = fa[node[i]], cid = 0;
while(x != nfa[node[i]]){
tres[que[node[i]][cid]] = x;
x = fa[x]; ++cid;
}
}
for(int i = 1; i <= tot; ++i)tres[nid[node[i]]] = node[i];
for(int i = 1; i <= n; ++i){
tr[bl][i] = tres[S.fa(i)];
if(!tr[bl][i])tr[bl][i] = kfa(i, R[bl] - L[bl] + 1);
}
for(int i = 1; i <= n; ++i)tres[i] = 0;
for(int i = 1; i <= n; ++i)thq[i] = 0;
for(int i = 1; i <= tot; ++i)it.head[node[i]] = 0;
for(int i = 1; i <= tot; ++i)que[node[i]].clear();
}
int trans(int l, int r, int x){
int thel = bl[l], ther = bl[r];
if(thel == ther){
for(int i = l; i <= r; ++i)x = move(x, a[i]);
return x;
}
for(int i = l; i <= R[thel]; ++i)x = move(x, a[i]);
for(int i = thel + 1; i < ther; ++i)x = tr[i][x];
for(int i = L[ther]; i <= r; ++i)x = move(x, a[i]);
return x;
}
int main(){
freopen("b.in","r",stdin);
freopen("b.out","w",stdout);
n = read(), m = read(), q = read(); B = max((int)sqrt(m * 0.5), 1);
for(int i = 2; i <= n; ++i)t.add(fa[i] = read(), i);
for(int i = 1; i <= m; ++i)a[i] = read();
for(int i = 1; i <= m; ++i)bl[i] = (B + i - 1) / B;
for(int i = 1; i <= bl[m]; ++i)L[i] = (i - 1) * B + 1;
for(int i = 1; i <= bl[m]; ++i)R[i] = i * B;
R[bl[m]] = m; dfs1(1); dfs2(1, 1);
for(int i = 1; i <= bl[m]; ++i)get_tr(i);
for(int i = 1; i <= q; ++i){
int l = read(), r = read();
printf("%d\n",trans(l, r, read()));
}
return 0;
}
C.C
没改。
