[考试总结]noip模拟27
作者：互联网
这个牛半仙可真的是人生赢家
这波中二的名字又是让我心头一震，然后有发现没有 \(PDF\) 和大样例。
没有大样例就意味着不能打挂一处。
然而这是让我最头疼的事情。。。
\(100\) 多行的代码一般对我来说也就是不到 \(5\) 分钟就能打完，然后开始 \(50min\) 调试。。。
有亿点亏。。。
其实这样的联赛模拟题目使劲想一想还是可以想出正解的，然而脑袋里不知道为什么总是想要去打一个暴力。。。
确实该改一改这个毛病。
牛半仙的妹子图
一开始看这个题目的时候就觉得这个出题人真的是中二至极。
这就是我打挂这个题目的理由
这个题目实际上很像奶牛跨栏那个题目，就是要求到达一个点路径上的最小值的最大。
然而当时所有人似乎用的都是 \(floyed\) 高级算法。。。
然后这个题目就不知道用 \(dij\) 咋搞。。。
其实也差不多，我想的是可以把所有边压进 ￥priority_queue$ 当中，这样的话也能维护这个。
然而因为不是很经常使用 \(dijkstra\) 和 \(priority_queue\)，所以。。。
我压根不知道该 反着重载。
所以我是这么写的。。。
class xin_data
{
private:
friend bool operator < (xin_data x,xin_data y)
{return x.w < y.w;}
public:
int w,pos;
};
然后就 \(gg\) 了。
因为 \(w\leq600\) ，所以可以建立 权值数组。
然后就能跑了。
#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 int long long
namespace xin_io
{
#define scanf nb = scanf
#define debug cout<<"debug"<<endl
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<2,stdin),p1 == p2) ? EOF : *p1++
char buf[1<<20],*p1 = buf,*p2 = buf,output[100]; FILE *xinnb;int nb;typedef long long ll; typedef unsigned long long ull;
void openfile() {xinnb = freopen("t.txt","r",stdin);} void outfile() {xinnb = freopen("o.txt","w",stdout);}
inline int get()
{
register int s = 0,f = 1; register char ch = gc(); while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc();return s * f;
}
template<typename type>inline void write(type x,char out)
{
if(!x) return putchar('0'),putchar(out),void(); if(x < 0) putchar('-'),x = -x;
register int cnt = 0;while(x) output[++cnt] = x % 10,x /= 10;
throw(i,cnt,1) putchar(output[i] xor 48);return putchar(out),void();
}
}
using namespace xin_io; static const int maxn = 1e6+10,inf = 1e9+7; static const ll llinf = 1e17+7;
namespace xin
{
class xin_edge{public:int next,ver,w;}edge[maxn];
int head[maxn],zhi = 0;
inline void add(int x,int y,int z) {edge[++zhi].ver = y;edge[zhi].w = z; edge[zhi].next = head[x]; head[x] = zhi;}
class xin_data
{
private:
friend bool operator < (xin_data x,xin_data y)
{return x.val > y.val;}
public:
int val,pos;
xin_data(){}
xin_data(int val,int pos):val(val),pos(pos){}
};
std::priority_queue<xin_data>q;
int n,m,qnum,st,op,mod,c[maxn],dis[maxn];
int ton[maxn],sum = 0;
int he[maxn],base[maxn],cnt,num[maxn];
bool vis[maxn];
inline short main()
{
#ifndef ONLINE_JUDGE
openfile();
#endif
n = get(); m = get(); qnum = get(); st = get(); op = get(); if(op) mod = get();
try(i,1,n) c[i] = get();
try(i,1,m)
{
register int x = get(),y = get(),z = get();
add(x,y,z); add(y,x,z);
}
ton[c[st]]++; sum ++;
for(register int i=head[st];i;i=edge[i].next) q.push(xin_data(edge[i].w,edge[i].ver));
int maxx = 0; base[++cnt] = 1; num[cnt] = sum;
while(!q.empty())
{
register int dis = q.top().val,now = q.top().pos;q.pop();
if(vis[now])continue;
vis[now] = 1; maxx = std::max(maxx,dis);
ton[c[now]]++; if(ton[c[now]] == 1) {sum++; if(base[cnt] xor maxx) base[++cnt] = maxx; num[cnt] = sum;}
for(register int i=head[now];i;i=edge[i].next) q.push(xin_data(edge[i].w,edge[i].ver));
}
base[++cnt] = inf;
try(i,1,cnt) he[i] = he[i-1] + num[i - 1] * (base[i] - base[i-1]);
int pre = 0;
try(i,1,qnum)
{
register int l = get(),r = get();
if(op) l = (l xor pre) % mod + 1,r = (r xor pre) % mod + 1;
if(r < l) std::swap(l,r);
register int lb = std::upper_bound(base+1,base+cnt+1,l) - base,rb = std::upper_bound(base+1,base+cnt+1,r) - base - 1;
register int ans = he[rb] - he[lb];
ans += (base[lb] - l) * num[lb - 1]; ans += (r - base[rb] + 1) * num[rb];
write(pre = ans,'\n');
}
return 0;
}
}
signed main(){return xin::main();}
牛半仙的妹子Tree
这个题目说实话我是真的没有想到暴力直接可过的。
然后加强数据之后。。。还是能过。。。
其实就是记录下来每一个 \(1\) 操作。
然后在每一个 \(3\) 操作的时候遍历整个 \(1\) 操作的数组。
然后而操作就是把那个数组的指针清零就行了。
然而 \(lca\) 一定要 \(\mathcal O(1)\) 求，不然一定会 \(T\)。
#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)
namespace xin_io
{
#define scanf nb = scanf
#define debug cout<<"debug"<<endl
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<2,stdin),p1 == p2) ? EOF : *p1++
char buf[1<<20],*p1 = buf,*p2 = buf,output[100]; FILE *xinnb;int nb;typedef long long ll; typedef unsigned long long ull;
void openfile() {xinnb = freopen("t.txt","r",stdin);} void outfile() {xinnb = freopen("o.txt","w",stdout);}
inline int get()
{
register int s = 0,f = 1; register char ch = gc(); while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc();return s * f;
}
template<typename type>inline void write(type x,char out)
{
if(!x) return putchar('0'),putchar(out),void(); if(x < 0) putchar('-'),x = -x;
register int cnt = 0;while(x) output[++cnt] = x % 10,x /= 10;
throw(i,cnt,1) putchar(output[i] xor 48);return putchar(out),void();
}
}
using namespace xin_io; static const int maxn = 1e6+10,inf = 1e9+7; static const ll llinf = 1e17+7;
namespace xin
{
class xin_edge{public:int next,ver;} edge[maxn];
int head[maxn],cnt = 0;
inline void add(int x,int y){edge[++cnt].ver = y; edge[cnt].next = head[x]; head[x] = cnt;}
int dep[maxn],f[maxn][32],lg[maxn],fr[maxn],num;
void dfs(int x,int fa)
{
dep[x]=dep[fa]+1;
f[++num][0] = x;
if(!fr[x])fr[x]=num;
for(int i=head[x];i;i=edge[i].next)
{
int dd=edge[i].ver;
if(dd==fa)continue;
dfs(dd,x);
f[++num][0]=x;
if(!fr[x])fr[x]=num;
}
}
void RMQ()
{
for(register int j=1;(1<<j)<=num;++j)
for(register int i=1;i+(1<<j)-1<=num;++i)
if(dep[f[i][j-1]]<dep[f[i+(1<<(j-1))][j-1]])f[i][j]=f[i][j-1];
else f[i][j]=f[i+(1<<(j-1))][j-1];
try(i,2,num) lg[i] = lg[i>>1] + 1;
}
int lca(int x,int y)
{
x = fr[x]; y = fr[y];
if(x > y) x ^= y ^= x ^= y;
int len=lg[y-x+1];
return dep[f[x][len]]<dep[f[y-(1<<len)+1][len]]?f[x][len]:f[y-(1<<len)+1][len];
}
inline int distance(int x,int y){return dep[x] + dep[y] - 2 * dep[lca(x,y)];}
int a[maxn],temp[maxn],zhi = 0;
int n,m;
bool cold[maxn];
void kuosan()
{
try(i,1,zhi) temp[i] = a[i];
int ms = zhi; zhi = 0;
try(i,1,ms)
{
register int x = temp[i];
for(register int i=head[x];i;i=edge[i].next)
{
register int y = edge[i].ver;
if(!cold[y]) cold[y] = 1,a[++zhi] = y;
}
}
}
class xin_data
{
public:
int tim,pos;
xin_data(){}
xin_data(int tim,int pos):tim(tim),pos(pos){}
}d[maxn];
bool sp1 = 1,sp2 = 1;
inline short main()
{
#ifndef ONLINE_JUDGE
openfile();
#endif
n = get(); m = get();
try(i,1,n-1)
{
register int x = get(),y = get();
add(x,y); add(y,x);
if(x xor 1) sp1 = 0;
if((x xor i) or (y xor (i + 1))) sp2 = 0;
}
if(sp1)
{
int tim = 0,cnt = 0;
try(i,1,m)
{
register int op = get(),x = get();
if(cnt)
{
if(cold[1]) cnt = n;
else cold[1] = 1;
}
if(op == 1)
{
cold[x] = 1;
cnt++;
}
if(op == 2)
{
cnt = 0;
memset(cold,0,sizeof(bool) * (n + 1));
}
if(op == 3)
{
if(cnt == n) printf("wrxcsd\n");
else printf(cold[x] ? "wrxcsd\n" : "orzFsYo\n");
}
}
return 0;
}
if(sp2)
{
try(i,1,m)
{
register int op = get(),x = get();
kuosan();
if(op == 1)
{
a[++zhi] = x;
cold[x] = true;
}
if(op == 2)
{
memset(cold,0,sizeof(bool) * (n + 1));
zhi = 0;
}
if(op == 3)
printf(cold[x] ? "wrxcsd\n" : "orzFsYo\n");
}
return 0;
}
dfs(1,0); RMQ();
try(tim,1,m)
{
register int op = get(),x = get();
if(op == 1)
{
d[++zhi] = xin_data(tim,x);
}
if(op == 2)
{
zhi = 0;
}
if(op == 3)
{
bool ok = 1;
try(i,1,zhi)
{
register int ks = tim - d[i].tim,dis = distance(x,d[i].pos);
if(ks >= dis) {ok = 0;break;}
}
puts(ok ? "orzFsYo" : "wrxcsd");
}
}
return 0;
}
}
signed main(){return xin::main();}
牛半仙的妹子序列
似乎是极长上升子序列的板子。
然而我不会。。。
然后 \(gg\) 了，直接这个题目抱 \(\color{red}{0}\) 了。
用线段树维护这个玩意，似乎和昨天的第二题有些相似。
#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)
namespace xin_io
{
#define scanf nb = scanf
#define debug cout<<"debug"<<endl
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<2,stdin),p1 == p2) ? EOF : *p1++
char buf[1<<20],*p1 = buf,*p2 = buf,output[100]; FILE *xinnb;int nb;typedef long long ll; typedef unsigned long long ull;
void openfile() {xinnb = freopen("t.txt","r",stdin);} void outfile() {xinnb = freopen("o.txt","w",stdout);}
inline int get()
{
register int s = 0,f = 1; register char ch = gc(); while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc();return s * f;
}
template<typename type>inline void write(type x,char out)
{
if(!x) return putchar('0'),putchar(out),void(); if(x < 0) putchar('-'),x = -x;
register int cnt = 0;while(x) output[++cnt] = x % 10,x /= 10;
throw(i,cnt,1) putchar(output[i] xor 48);return putchar(out),void();
}
}
using namespace xin_io; static const int maxn = 1e6+10,inf = 1e9+7,mod = 998244353; static const ll llinf = 1e17+7;
namespace xin
{
int n;
int c[maxn],pos[maxn];
int f[maxn],maxx;
class xin_segment
{
private:
#define ls(fa) (fa << 1)
#define rs(fa) (fa << 1 | 1)
int lim(int fa,int val)
{
if(t[fa].tag) return t[fa].s > val ? f[t[fa].s] : 0;
return t[rs(fa)].s < val ? lim(ls(fa),val) : (t[fa].lans + lim(rs(fa),val)) % mod;
}
inline void up(int fa)
{
t[fa].s = std::max(t[ls(fa)].s,t[rs(fa)].s);
t[fa].lans = lim(ls(fa),t[rs(fa)].s);
}
public:
class xin_tree{public:int s,lans,tag;}t[maxn];
void build(int fa,int l,int r)
{
if(l == r) return t[fa].tag = 1,void(); //tag the leave point
register int mid = l + r >> 1;
build(ls(fa),l,mid); build(rs(fa),mid+1,r);
}
void modify(int fa,int l,int r,int pos,int val)
{
if(pos > r or pos < l) return;
if(l == r) return t[fa].s = val,void();
register int mid = l + r >> 1;
modify(ls(fa),l,mid,pos,val); modify(rs(fa),mid+1,r,pos,val);
up(fa);
}
int query(int fa,int l,int r,int ql,int qr)
{
if(ql > r or qr < l) return 0;
if(ql <= l and qr >= r)
{
register int temp = maxx; maxx = std::max(maxx,t[fa].s);
return lim(fa,temp);
}
register int mid = l + r >> 1,res = 0;
res = (res + query(rs(fa),mid+1,r,ql,qr)) % mod; res = (res + query(ls(fa),l,mid,ql,qr)) % mod;
return res;
}
}t;
class xin_bit
{
private:
#define lowbit(x) (x & -x)
public:
int s[maxn];
inline void add(int x,int val)
{for(register int i=x;i<=n;i+=lowbit(i)) s[i] += val;}
inline int ask(int x)
{int res = 0;for(register int i=x;i;i-=lowbit(i)) res += s[i];return res;}
}bit;
inline short main()
{
#ifndef ONLINE_JUDGE
openfile();
#endif
n = get(); try(i,1,n) c[i] = get(),pos[c[i]] = i;
t.build(1,1,n);
try(i,1,n)
{
register int x = pos[i]; maxx = 0;
f[i] = t.query(1,1,n,1,x);
if(!f[i]) f[i] = 1;
t.modify(1,1,n,x,i);
}
int ans = 0;
throw(i,n,1)
{
if(bit.ask(n) == bit.ask(pos[i])) ans = (ans + f[i]) % mod;
bit.add(pos[i],1);
}
cout<<ans<<endl;
return 0;
}
}
signed main(){return xin::main();}
标签：cnt,27,return,noip,int,xin,fa,maxn,模拟 来源： https://www.cnblogs.com/NP2Z/p/15084779.html