[考试总结]noip模拟16
达成成就，一天更3篇总结。
又是一个暴力场
别问我为什么开局 \(5\) 分钟就问老师为什么 \(T3\) 没有提交的窗口。
开题读题，一路自闭到 \(T3\) ，发现 \(T3\) 可打暴力，所以一波暴力打上去，开始只有 \(30pts\)
然后转向 \(T2\) ，发现也有部分分数可以打，所以就开始打这个题目的暴力。
瞄准 \(20pts\) 部分分数，然后疯狂开始打。
打完之后也不能放弃啊，所以开始回去开 \(T1\)，然后开始自闭。
给的样例手摸都出不来。。。。
自闭\(*2\)
突然思路开始涌现，发现可以把坐标系分成 \(1000\) * \(1000\)，然后开始一个一个点枚举，这样的话就能骗过\(Special\;Judge\)
然后开始打。。。
然后样例都调不出来。。。
自闭\(*3\)
但是经过我的奋战，\(T1\) 终于出来了样例。。。
然后转向 \(T3\)，感觉数组好像开的有点小，但是部分分数似乎也够，但还好我开打了一些。又赚了 \(20pts\)
不扯淡了
T1:
首先考虑的应该是二分做法。
然后如果打对就可以得到 \(80pts\)
然而只有土哥打对了。。。。
战神看出了正解最小生成树，然而 \(Kruskal\) 爆炸了。。。
然而还是极其强大。。。
正解就是最小生成树，然后 \(Kruskal\) 也可以过。
就是每个点连边，然后和上下边界连边。
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define debug cout<<"debug"<<endl
//#define int long long
namespace xin_io
{
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
#define scanf eat1 = scanf
#define freopen eat2 = freopen
int eat1; FILE *eat2; char buf[1<<20],*p1 = buf,*p2 = buf;
inline void openfile() {freopen("t.txt","r",stdin);} inline void outfile() {freopen("o.txt","w",stdout);}
template<class type>inline type get()
{
type s = 0,f = 1; register char ch = gc();
while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
while(isdigit(ch)) {s = s * 10 + ch - '0'; ch = gc();}
return s * f;
}
}
using namespace xin_io; static const int maxn = 1e6+10,inf = 0x7f7f7f,mod = (1<<30);
#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)
typedef long long ll;
namespace xin
{
const double eps = 1e-8;
int n,m,k;
class xin_data{public:int x,y;}d[maxn];
inline double getdis(double x1,double y1,double x2,double y2)
{return std::sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));}
inline bool equal(double x,double y)
{
if(abs(x - y) <= eps) return true;
return false;
}
int x[maxn],y[maxn];
double dis[maxn];
bool vis[maxn];
double ans = -inf * 1.0;
inline short main()
{
#ifndef ONLINE_JUDGE
openfile();
#endif
n = get<signed>(); m = get<signed>(); k = get<signed>();
try(i,1,k) x[i] = get<signed>(),y[i] = get<signed>(),dis[i] = (double)(m - y[i]);
dis[k+1] = m; dis[0] = inf * 1.0;
while(true)
{
register int temp = 0;
try(i,1,k+1)if(!vis[i] and dis[i] < dis[temp]) temp = i;
vis[temp] = true;
ans = std::max(ans,dis[temp]);
if(temp == k + 1) break;
try(i,1,k) dis[i] = std::min(dis[i],getdis(x[i],y[i],x[temp],y[temp]));
dis[k+1] = std::min(dis[k+1],y[temp] * 1.0);
}
printf("%0.9lf\n",ans/2.0);
return 0;
}
}
signed main() {return xin::main();}
T2:
要找一个极长上升序列。
然后线段树优化。。。。
老猥琐了。。
自己研究吧。。。
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define debug cout<<"debug"<<endl
//#define int long long
namespace xin_io
{
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
#define scanf eat1 = scanf
#define freopen eat2 = freopen
int eat1; FILE *eat2; char buf[1<<20],*p1 = buf,*p2 = buf;
inline void openfile() {freopen("t.txt","r",stdin);} inline void outfile() {freopen("o.txt","w",stdout);}
template<class type>inline type get()
{
type s = 0,f = 1; register char ch = gc();
while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
while(isdigit(ch)) {s = s * 10 + ch - '0'; ch = gc();}
return s * f;
}
}
using namespace xin_io; static const int maxn = 1e6+10,inf = 0x7f7f7f7f,mod = (1<<30);
#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)
typedef long long ll;
namespace xin
{
int p[maxn],c[maxn];
int n;
int f[maxn];
int maxx = 0;
class xin_segment
{
private:
#define ls(fa) (fa << 1)
#define rs(fa) (fa << 1 | 1)
inline void up(int fa,int l,int r,int x)
{
register int mid = l + r >> 1;
t[fa].mf = find(ls(fa),l,mid,l,mid,t[rs(fa)].mx);
t[fa].f = std::min(t[ls(fa)].f,t[rs(fa)].f);
t[fa].mx = std::max(t[ls(fa)].mx,t[rs(fa)].mx);
// cout<<"t[fa].mx = "<<t[fa].mx<<endl;
}
public:
class xin_tree{public:int mx,mf,f;}t[maxn];
int find(int fa,int l,int r,int ql,int qr,int p)
{
if(t[fa].mx < p) return inf;
if(l == r) {maxx = std::max(maxx,t[fa].mx); return t[fa].f;}
register int mid = l + r >> 1;
if(l == ql and r == qr)
{
if(t[rs(fa)].mx < p) return find(ls(fa),l,mid,ql,mid,p); maxx = std::max(maxx,t[fa].mx); //cout<<"maxx = "<<maxx<<endl;
return std::min(t[fa].mf,find(rs(fa),mid+1,r,mid+1,qr,p));
}
if(qr <= mid) return find(ls(fa),l,mid,ql,qr,p);
else if(ql > mid) return find(rs(fa),mid+1,r,ql,qr,p);
else {maxx = 0; register int temp = find(rs(fa),mid+1,r,mid+1,qr,p);return std::min(temp,find(ls(fa),l,mid,ql,mid,maxx));}
}
void insert(int fa,int l,int r,int x,int p)
{
if(x > r or x < l) return ;
if(l == r) {t[fa].f = f[p]; t[fa].mx = p; return;}
register int mid = l + r >> 1;
insert(ls(fa),l,mid,x,p); insert(rs(fa),mid+1,r,x,p);
up(fa,l,r,x);
}
}t;
inline short main()
{
#ifndef ONLINE_JUDGE
openfile();
#endif
n = get<signed>();
try(i,1,n) p[i] = get<signed>();
try(i,1,n) c[i] = get<signed>();
try(i,1,(n<<2)) t.t[i].f = t.t[i].mf = inf;
for(register int i=1;i<=(n<<2);i<<=1) t.t[i].f = t.t[i].mf = 0;
try(i,1,n)
{
f[i] = t.find(1,0,n,0,p[i] - 1,0) + c[i];
t.insert(1,0,n,p[i],i);
}
register int ans = inf,temp = 0;
throw(i,n,1) if(temp < p[i]) temp = p[i],ans = std::min(ans,f[i]);
cout<<ans<<endl;
return 0;
}
}
signed main() {return xin::main();}
T3:
咱们把式子变换一下
然后发现：
\[\frac {c_u-c_v}{dis_{v,u}}=\frac {c_u-c_v}{d_v-d_u}=-\frac {c_v-c_u} {d_v-d_u}
\]
然后就是一个斜率式子。
然后维护凸包，找到斜率最大的。
采用倍增的方法，然后飞快。
防止黑心出题人
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define debug cout<<"debug"<<endl
//#define int long long
namespace xin_io
{
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
#define scanf eat1 = scanf
#define freopen eat2 = freopen
int eat1; FILE *eat2; char buf[1<<20],*p1 = buf,*p2 = buf;
inline void openfile() {freopen("t.txt","r",stdin);} inline void outfile() {freopen("o.txt","w",stdout);}
template<class type>inline type get()
{
type s = 0,f = 1; register char ch = gc();
while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
while(isdigit(ch)) {s = s * 10 + ch - '0'; ch = gc();}
return s * f;
}
}
using namespace xin_io; static const int maxn = 1e6+10,inf = 0x7f7f7f,mod = (1<<30);
#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)
typedef long long ll;
namespace xin
{
class xin_edge{public:int next,ver;}edge[maxn];
int d[maxn],fa[maxn][31],ans[maxn],c[maxn],n;
int head[maxn],zhi = 0;
inline void add(int x,int y) {edge[++zhi].ver = y; edge[zhi].next = head[x]; head[x] = zhi;}
inline double getk(int x,int y) {return (double)(c[x] - c[y]) / (double)(d[y] - d[x]);}
void dfs(int x)
{
register int now = fa[x][0];
throw(i,30,0)
{
if(fa[now][i] <= 1) continue;
if(getk(x,fa[now][i]) >= getk(x,fa[fa[now][i]][0])) now = fa[fa[now][i]][0]; //just jump to the best
}
if(now > 1 and getk(x,now) >= getk(x,fa[now][0])) now = fa[now][0];
ans[x] = now; fa[x][0] = now; //this method is not totally the same as before,f[x][0] is its best answer but not his father
try(i,1,30) fa[x][i] = fa[fa[x][i-1]][i-1];
for(register int i=head[x];i;i=edge[i].next)
{
register int y = edge[i].ver;
d[y] = d[x] + 1; dfs(y);
}
}
inline short main()
{
#ifndef ONLINE_JUDGE
openfile();
#endif
n = get<signed>();
try(i,1,n) c[i] = get<signed>();
try(i,2,n)
{
fa[i][0] = get<signed>();
add(fa[i][0],i);
}
d[1] = 1;
dfs(1);
try(i,2,n) printf("%.10lf\n",getk(i,ans[i]));
return 0;
}
}
signed main() {return xin::main();}
