UOJ 2021 NOI Day2 部分题解
获奖名单
Solution
不难看出，若我们单个 \(x\) 连 \((0,x),(x,0)\)，两个连 \((x,y),(y,x)\) ，除去中间过对称轴的一个两个组，就是找很多个欧拉回路。
直接来就好了。
Code
#include <bits/stdc++.h>
using namespace std;
#define Int register int
#define MAXN 500005
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
#define pii pair<int,int>
#define se second
#define fi first
int n,m;
bool vis[MAXN];
vector <pii> g[MAXN];
int cnt,seq[MAXN];
void dfs (int u){
while (!g[u].empty()){
pii it = g[u].back();
g[u].pop_back();
if (vis[abs(it.se)]) continue;
vis[abs (it.se)] = 1,dfs (abs (it.fi)),seq[++ cnt] = it.se;
}
}
signed main(){
read (n,m);
for (Int i = 1;i <= n;++ i){
int opt,x,y;read (opt,x);
if (opt == 1) g[0].push_back ({x,-i}),g[x].push_back ({0,i});
else read (y),g[x].push_back ({y,i}),g[y].push_back ({x,-i});
}
dfs (0);
vector <int> ansl,ansr;
for (Int i = 1;i <= cnt;i += 2) ansl.push_back (seq[i]);
for (Int i = 2;i <= cnt;i += 2) ansr.push_back (-seq[i]);
int mid = 0;
for (Int i = 1;i <= m;++ i){
sort (g[i].begin(),g[i].end());
for (Int j = 0;j < g[i].size();j += 2){
if (g[i][j].fi == i && g[i][j].se == -g[i][j + 1].se){
mid = abs(g[i][j].se);
continue;
}
if (g[i][j].fi > i || (g[i][j].fi == i && g[i][j].se < 0))
ansl.push_back (g[i][j].se),
ansr.push_back (-g[i][j + 1].se);
}
}
for (Int i = ansl.size() - 1;~i;-- i) write (abs(ansl[i])),putchar (' ');
if (mid) write (mid),putchar (' ');
for (Int i = 0;i < ansr.size();++ i) write (abs(ansr[i])),putchar (' ');
putchar ('\n');
for (Int i = ansl.size() - 1;~i;-- i) write (ansl[i] < 0),putchar (' ');
if (mid) write (0),putchar (' ');
for (Int i = 0;i < ansr.size();++ i) write (ansr[i] < 0),putchar (' ');
putchar ('\n');
return 0;
}
诡异操作
Solution
考虑使用线段树维护，对于一个线段树上的区间，我们可以维护 \(s_c\) 表示二进制位出现次数二进制第 \(c\) 位为 \(1\) 的和。你发现这个东西可以进行合并，并且取并操作可以直接对于每一个 \(s_c\) 并。
然后对于下取整操作可以直接暴力重构。
复杂度是个米奇妙妙复杂度，据说是 \(\Theta(128n+q\log^2n)\)。
Code
#include <bits/stdc++.h>
using namespace std;
#define Int register int
#define MAXN 300005
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
typedef __uint128_t u128;
inline u128 read() {
static char buf[100];
scanf("%s", buf);
// std::cin >> buf;
u128 res = 0;
for(int i = 0;buf[i];++i) {
res = res << 4 | (buf[i] <= '9' ? buf[i] - '0' : buf[i] - 'a' + 10);
}
return res;
}
inline void output(u128 res) {
if(res >= 16)
output(res / 16);
putchar(res % 16 >= 10 ? 'a' + res % 16 - 10 : '0' + res % 16);
//std::cout.put(res % 16 >= 10 ? 'a' + res % 16 - 10 : '0' + res % 16);
}
#define ll u128
ll a[MAXN];
int n,q,len[MAXN << 2];
ll res = ~ll(0);
struct Segment{
ll val[MAXN << 2][20],any[MAXN << 2],laz[MAXN << 2];
void pushup (int x){
memset (val[x],0,sizeof (u128) * len[x]);
int ls = x << 1,rs = x << 1 | 1;
val[x][0] = val[ls][0] ^ val[rs][0];ll car = val[ls][0] & val[rs][0];
for (Int i = 1;i < len[x];++ i){
ll w = val[ls][i] ^ val[rs][i];
val[x][i] = w ^ car,car = (car & w) | (val[ls][i] & val[rs][i]);
}
val[x][len[x]] = car,any[x] = any[x << 1] | any[x << 1 | 1];
}
ll getv (int x){
ll ans = 0;
for (Int i = 0;i <= len[x];++ i) ans += val[x][i] << i;
return ans;
}
void pushadd (int x,ll V){
for (Int i = 0;i <= len[x];++ i) val[x][i] &= V;
laz[x] &= V,any[x] &= V;
}
void pushdown (int x){pushadd (x << 1,laz[x]),pushadd (x << 1 | 1,laz[x]),laz[x] = res;}
void build (int x,int l,int r){
laz[x] = res,len[x] = 1;
while ((1 << len[x]) <= (r - l + 1)) ++ len[x];
if (l == r) return val[x][0] = any[x] = a[l],void ();
int mid = l + r >> 1;
build (x << 1,l,mid),build (x << 1 | 1,mid + 1,r);
pushup (x);
}
void modify1 (int x,int l,int r,int ql,int qr,ll V){
if (l >= ql && r <= qr) return pushadd (x,V),void ();
int mid = l + r >> 1;pushdown (x);
if (ql <= mid) modify1 (x << 1,l,mid,ql,qr,V);
if (qr > mid) modify1 (x << 1 | 1,mid + 1,r,ql,qr,V);
pushup (x);
}
void div (int x,int l,int r,ll V){
if (!any[x]) return ;
if (l == r) return any[x] = (val[x][0] /= V),void ();
int mid = l + r >> 1;pushdown (x);
div (x << 1,l,mid,V),div (x << 1 | 1,mid + 1,r,V);
pushup (x);
}
ll query (int x,int l,int r,int ql,int qr){
if (l >= ql && r <= qr) return getv (x);
int mid = l + r >> 1;pushdown (x);ll res = 0;
if (ql <= mid) res += query (x << 1,l,mid,ql,qr);
if (qr > mid) res += query (x << 1 | 1,mid + 1,r,ql,qr);
return res;
}
void modify2 (int x,int l,int r,int ql,int qr,ll V){
if (l >= ql && r <= qr) return div (x,l,r,V);
int mid = l + r >> 1;pushdown (x);
if (ql <= mid) modify2 (x << 1,l,mid,ql,qr,V);
if (qr > mid) modify2 (x << 1 | 1,mid + 1,r,ql,qr,V);
pushup (x);
}
}T;
signed main(){
read (n,q);
for (Int i = 1;i <= n;++ i) a[i] = read ();
T.build (1,1,n);
while (q --> 0){
int opt,qL,qR;ll V;
read (opt,qL,qR);
if (opt <= 2){
V = read ();
if (opt == 1 && V > 1) T.modify2 (1,1,n,qL,qR,V);
else if (opt == 2) T.modify1 (1,1,n,qL,qR,V);
}
else output (T.query (1,1,n,qL,qR)),putchar ('\n');
}
return 0;
}
