标签：noip int max register 50 mid try maxn 模拟
第零题
发现一个结论，就是说这个道路怎么走答案其实是一样的。
然后就能愉快暴力了。。。
话说为何班长每次暴力都能 \(Ac\)
但是我们要倍增！！！
然后就行了。。。
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register signed i=a;i<=b;++i)
#define throw(i,a,b) for(register signed i=a;i>=b;--i)
#define go(i,x) for(register signed i=head[x],y=edge[i].ver;i;i=edge[i].next,y=edge[i].ver)
namespace xin_io
{
#define sb(x) cout<<#x" = "<<x<<' '
#define jb(x) cout<<#x" = "<<x<<endl
#define debug cout<<"debug"<<endl
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
char buf[1<<20],*p1 = buf,*p2 = buf; int ak; typedef long long ll; typedef unsigned long long ull;
class xin_stream{public:template<typename type>inline xin_stream &operator >> (type &s)
{
register int f = 0;s = 0; register char ch = gc();
while(!isdigit(ch)) {f |= ch == '-'; ch = gc();}
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc(); return s = f ? -s : s,*this;
}}io;
}
using namespace xin_io; static const int maxn = 1e6+10,inf = 1e9+7; const ll llinf = 1e18+7;
#define int long long
namespace xin
{
class xin_edge{public:int next,ver,w;}edge[maxn];
int head[maxn],rp;
auto add = [](int x,int y,int z) -> void {edge[++rp].ver = y; edge[rp].w = z; edge[rp].next = head[x]; head[x] = rp;};
int siz[maxn],d[maxn],fa[maxn],top[maxn],hson[maxn],mi[23],he[maxn],c[maxn],sum[maxn],f[maxn][23],ton[maxn];
int n,k,m,tot,logms;
void dfs1(int x,int f)
{
d[x] = d[f] + 1; siz[x] = 1; fa[x] = f;
go(i,x)
{
if(y == f) continue;
c[y] = edge[i].w; dfs1(y,x); siz[x] += siz[y];
if(siz[y] > siz[hson[x]]) hson[x] = y;
}
}
void dfs2(int x,int t)
{
top[x] = t; if(hson[x]) dfs2(hson[x],t);
go(i,x)
{
if(y == hson[x] or y == fa[x]) continue;
dfs2(y,y);
}
}
void dfs(int x)
{
auto find = []() -> int
{
register int lim = sum[tot] + k,l = 0,r = tot,ret = 0;
while(l <= r)
{
register int mid = l + r >> 1;
if(sum[mid] >= lim) ret = mid,l = mid + 1;
else r = mid - 1;
}
return ton[ret];
};
ton[++tot] = x; sum[tot] = sum[tot - 1] - c[x];
f[x][0] = find();// jb(f[x][0]);
try(i,1,logms) f[x][i] = f[f[x][i-1]][i-1];
go(i,x)
{
if(y == fa[x]) continue;
he[y] = he[x] + c[y];
dfs(y);
}
tot --;
}
inline short main()
{
io >> n >> k; logms = log2(n) + 1ll;
// jb(logms);
try(i,1,n-1)
{
register int x,y,z; io >> x >> y >> z;
add(x,y,z); add(y,x,z);
}
auto prework = [=]() -> void {mi[0] = 1; try(i,1,logms) mi[i] = mi[i-1] << 1;sum[0] = llinf;}; prework();
dfs1(1,0); dfs2(1,1);
dfs(1);
// try(i,1,n) try(j,0,logms) sb(i),sb(j),jb(f[i][j]);
auto query = [=](int s,int t)mutable -> int
{
auto lca = [s,t]()mutable -> int
{
while(top[s] xor top[t])
{
if(d[top[s]] < d[top[t]]) s ^= t ^= s ^= t;
s = fa[top[s]];
}
return d[s] > d[t] ? t : s;
};
register int ans = 0,allca = lca();
throw(i,logms,0) if(d[f[s][i]] >= d[allca]) s = f[s][i],ans += mi[i];
throw(i,logms,0) if(d[f[t][i]] >= d[allca]) t = f[t][i],ans += mi[i];
if(he[t] - he[allca] >= k - (he[s] - he[allca])) return ans + 1; return ans;
};
io >> m;
try(i,1,m){register int s,t; io >> s >> t;printf("%lld\n",query(s,t));}
return 0;
}
}
signed main() {return xin::main();}
第负一题
首先想到垃圾基础 \(dp\)
就是：
\[f_{i,0} = max(f_{i-1,0},f_{i-1,1})\\ f_{i,1} = max(f_{i-2,1},f_{i-1,0}) \]然后愉快 \(\mathcal O(n^2)\)
然后这就是暴力。。
但是为啥只给 \(20pts\)
靠。。。。
然后正解我们加一个分治。
从中间断开。
然后 \(f_{i,1/0,1/0}\) 表示第 \(i\) 个，然后 \(mid\) 能不能选，然后是选了没选。
转移也很简单。。
l[i][1][1] = l[i+1][1][0] + a[i]; l[i][1][0] = max(l[i+1][1][1],l[i+1][1][0]);
l[i][0][1] = l[i+1][0][0] + a[i]; l[i][0][0] = max(l[i+1][0][1],l[i+1][0][0]);
l[i][1][1] = max(l[i][1][1],l[i][1][0]),l[i][0][1] = max(l[i][0][0],l[i][0][1]);
r[i][1][1] = r[i-1][1][0] + a[i]; r[i][1][0] = max(r[i-1][1][1],r[i-1][1][0]);
r[i][0][1] = r[i-1][0][0] + a[i]; r[i][0][0] = max(r[i-1][0][1],r[i-1][0][0]);
r[i][1][1] = max(r[i][1][1],r[i][1][0]),r[i][0][1] = max(r[i][0][0],r[i][0][1]);
然后代码：
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register signed i=a;i<=b;++i)
#define throw(i,a,b) for(register signed i=a;i>=b;--i)
#define go(i,x) for(register signed i=head[x],y=edge[i].ver;i;i=edge[i].next,y=edge[i].ver)
namespace xin_io
{
#define sb(x) cout<<#x" = "<<x<<' '
#define jb(x) cout<<#x" = "<<x<<endl
#define debug cout<<"debug"<<endl
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
char buf[1<<20],*p1 = buf,*p2 = buf; int ak; typedef long long ll; typedef unsigned long long ull;
class xin_stream{public:template<typename type>inline xin_stream &operator >> (type &s)
{
register int f = 0;s = 0; register char ch = gc();
while(!isdigit(ch)) {f |= ch == '-'; ch = gc();}
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc(); return s = f ? -s : s,*this;
}}io;
}
using namespace xin_io; static const int maxn = 1e6+10,inf = 1e9+7; const ll llinf = 1e18+7;
#define int long long
namespace xin
{
const int mod = 998244353;
int a[maxn],l[maxn][2][2],r[maxn][2][2],tl[maxn],tr[maxn];
int n,ans = 0;
auto max = [](int x,int y) -> int {return x > y ? x : y;};
void dfs(int ql,int qr)
{
if(qr - ql == 0) return (ans += a[qr]) %= mod,void();
if(qr - ql == 1) return (ans += max(a[ql],a[qr]) + a[ql] + a[qr]) %= mod,void();
register int mid = ql + qr >> 1;
dfs(ql,mid); dfs(mid+1,qr);
l[mid][1][1] = a[mid]; r[mid+1][1][1] = a[mid+1];
l[mid][1][0] = l[mid][0][0] = r[mid+1][1][0] = r[mid+1][0][0] = 0;
throw(i,mid-1,ql)
{
l[i][1][1] = l[i+1][1][0] + a[i]; l[i][1][0] = max(l[i+1][1][1],l[i+1][1][0]);
l[i][0][1] = l[i+1][0][0] + a[i]; l[i][0][0] = max(l[i+1][0][1],l[i+1][0][0]);
}
throw(i,mid-1,ql)
l[i][1][1] = max(l[i][1][1],l[i][1][0]),l[i][0][1] = max(l[i][0][0],l[i][0][1]);
try(i,mid+2,qr)
{
r[i][1][1] = r[i-1][1][0] + a[i]; r[i][1][0] = max(r[i-1][1][1],r[i-1][1][0]);
r[i][0][1] = r[i-1][0][0] + a[i]; r[i][0][0] = max(r[i-1][0][1],r[i-1][0][0]);
}
try(i,mid+2,qr)
r[i][1][1] = max(r[i][1][1],r[i][1][0]),r[i][0][1] = max(r[i][0][0],r[i][0][1]);
try(i,ql,mid) tl[i] = max(l[i][1][1] - l[i][0][1],0);
try(i,mid+1,qr) tr[i] = max(r[i][1][1] - r[i][0][1],0);
std::sort(tl+ql,tl+mid+1); std::sort(tr+mid+1,tr+qr+1);
register int temp = 0;
try(i,ql,mid) (temp += l[i][0][1]) %= mod; (temp *= (qr - mid)) %= mod;
(ans += temp) %= mod; temp = 0;
try(i,mid+1,qr) (temp += r[i][0][1]) %= mod;
(ans += temp * (mid - ql + 1) % mod) %= mod;
register int zhi = mid;
try(i,ql,mid)
{
while(zhi < qr and tr[zhi+1] <= tl[i]) zhi ++;
(ans += (zhi - mid) * tl[i] % mod) %= mod;
}
zhi = ql-1;
try(i,mid+1,qr)
{
while(zhi < mid and tl[zhi+1] < tr[i]) zhi ++;
(ans += (zhi - ql + 1) * tr[i] % mod) %= mod;
}
}
inline short main()
{
io >> n;
try(i,1,n) io >> a[i];
dfs(1,n);
cout<<ans<<endl;
return 0;
}
}
signed main() {return xin::main();}
题目自己推。。。
我只会 \(90pts\) 算法。
就是一个时间一个时间的看。
然后统计，
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register signed i=a;i<=b;++i)
#define throw(i,a,b) for(register signed i=a;i>=b;--i)
#define go(i,x) for(register signed i=head[x],y=edge[i].ver;i;i=edge[i].next,y=edge[i].ver)
namespace xin_io
{
#define sb(x) cout<<#x" = "<<x<<' '
#define jb(x) cout<<#x" = "<<x<<endl
#define debug cout<<"debug"<<endl
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
char buf[1<<20],*p1 = buf,*p2 = buf; int ak; typedef long long ll;// typedef unsigned long long ull;
class xin_stream{public:template<typename type>inline xin_stream &operator >> (type &s)
{
register int f = 0;s = 0; register char ch = gc();
while(!isdigit(ch)) {f |= ch == '-'; ch = gc();}
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc(); return s = f ? -s : s,*this;
}}io;
}
using namespace xin_io; static const int maxn = 1e7+10,inf = 1e9+7; const ll llinf = 1e18+7;
#define int long long
namespace xin
{
const int mod = 998244353;
typedef unsigned long long u64;
auto ksm = [](int x,int y,int ret = 1) -> int
{
while(y)
{
if(y & 1) ret = ret * x % mod;
x = x * x % mod; y >>= 1;
}
return ret;
};
u64 xorshift128p(u64 &A, u64 &B) {
u64 T = A, S = B;
A = S;
T ^= T << 23;
T ^= T >> 17;
T ^= S ^ (S >> 26);
B = T;
return T + S;
}
void gen(int n, int L, int X, int Y, u64 A, u64 B, int l[], int r[]) {
for (int i = 1; i <= n; i ++) {
l[i] = xorshift128p(A, B) % L + X;
r[i] = xorshift128p(A, B) % L + Y;
if (l[i] > r[i]) std::swap(l[i], r[i]);
}
}
int l[maxn],r[maxn],nl[maxn],nr[maxn];
int n,L,X,Y;
u64 A,B;
int ans = 0;
bool vis[maxn];
int f[maxn];
int q1[maxn],q2[maxn],zhi1 = 0,zhi2 = 0;
int tim = 0,wan = 0;
inline short main()
{
io >> n >> L >> X >> Y >> A >> B;
gen(n,L,X,Y,A,B,l,r);
try(i,1,n) q1[++zhi1] = i;
// try(i,1,n) sb(l[i]),jb(r[i]);
while(1)
{
tim++;
while(zhi1)
{
register int x = q1[zhi1 --];
q2[++zhi2] = x;
auto max = [](int x,int y) -> int {return x > y ? x : y;};
auto min = [](int x,int y) -> int {return x < y ? x : y;};
register int newl = max(l[x],max(l[x-1],l[x+1])),newr = min(r[x],min(r[x+1],r[x-1]));
if(newl == l[x]) newl ++;
if(newr == r[x]) newr --;
// sb(newl); jb(newr);
if(newr < newl or !newr or !newl)
{
vis[x] = 1;
wan ++;
f[x] = tim;
nl[x] = nr[x] = 0;
}
else nl[x] = newl,nr[x] = newr;
}
while(zhi2)
{
register int x = q2[zhi2 --];
l[x] = nl[x]; r[x] = nr[x];
if(!vis[x]) q1[++zhi1] = x;
}
if(n == wan) break;
}
// try(i,1,n) sb(i),jb(f[i]);
try(i,1,n) (ans += ksm(3,i-1) * f[i] % mod) %= mod;
cout<<ans<<endl;
return 0;
}
}
signed main() {return xin::main();}
标签：noip,int,max,register,50,mid,try,maxn,模拟 来源： https://www.cnblogs.com/NP2Z/p/15270555.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。