话说比较简单.除了第三题不会写平衡树啊你妹!!边做边写吧.
仍然是机智的链接
第一题:
题目的数据范围非常小,来个 大法狮 吧.$6^6$只有$46656$,加上浮点运算的超大常数也不多嘛 ╮(╯_╰)╭
实现有点恶心,暴力你懂得.
#include <cstdio>
#include <cmath>
#define PI 3.1415926535897932384626
inline double ab(double x){
return (((x)>=0)?(x):(-x));
}
inline double mi(double x,double y){
y=ab(y);
return (x<y?x:y);
}
inline double ma(double x,double y){
y=ab(y);
return (x>y?x:y);
}
int n;
double x[8],y[8],r[8],s[8],minLeft;
bool f[8];
void dfs(int p,int d,double si){
double a,b,c;
int i;
b=ab(x[p]-x[0]);
b=mi(b,x[p]-x[1]);
b=mi(b,y[p]-y[1]);
b=mi(b,y[p]-y[0]);
for(i=2;i<=n;++i){
if(f[i]) b=mi(b,hypot(x[p]-x[i],y[p]-y[i])-r[i]);
}
f[p]=true;
si+=PI*b*b;
r[p]=b;
if(d==n-1){
si=s[0]-si;
if(si<minLeft){
minLeft=si;
}
r[p]=0.0;
f[p]=false;
return;
}
for(i=2;i<=n;++i){
if(!f[i]){
dfs(i,d+1,si);
}
}
r[p]=0.0;
f[p]=false;
}
int main(){
freopen("box2.in","r",stdin);
freopen("box2.out","w",stdout);
scanf("%d%lf%lf%lf%lf",&n,x,y,x+1,y+1);
minLeft=s[0]=s[1]=ab((x[1]-x[0])*(y[1]-y[0]));
++n;
int i;
for(i=2;i<=n;++i) scanf("%lf%lf",x+i,y+i);
for(i=2;i<=n;++i){
dfs(i,1,0.0);
}
printf("%.0lf",minLeft);
return 0;
}
~!~!@~!@@!@!##@@#@!@##@@@#@#%^&^%##@!@@我是机智的分割线~~~
第二题:
简单的动规.DP方程:
DP F[i,j]=max(F[i-1,j]+same(c[i],i-j),F[i-1,j-1])
# F[i,j]表示前i个数删去j个的最大匹配数,c[i]表示第i个数字
# same(a,b)=(a==b?1:0)
return MaxAmong(F,lengthof F)
#include <cstdio>
int f[2000][2000],c[2000],n,i,j;
inline int max(int a,int b){
return a>b?a:b;
}
int main(){
freopen("sequence.in","r",stdin);
freopen("sequence.out","w",stdout);
scanf("%d",&n);
for(i=1;i<=n;++i){
scanf("%d",c+i);
}
for(i=1;i<=n;++i){
for(j=0;j<=i;++j){
f[i][j]=max(f[i-1][j]+(c[i]==i-j),f[i-1][j-1]);
}
}
j=0;
for(i=0;i<=n;++i){
if(f[n][i]>j) j=f[n][i];
}
printf("%d
",j);
return 0;
}
真的很 短啊.
第三题:
不会写BST啊神犇滚粗!!!
好吧今天边学习边写,应该SBT比较方便实现吧.
妈旦谁骗我SBT好实现的???
贴代码:
#include <cstdio>
#include <algorithm>
struct node{
int key , size;
node *c[2];
node(int k){
key = k;
size = 1;
};
void t(){size=c[0]->size+c[1]->size+1;}
};
class SBT{
private:
node * root , * nil;
inline void insert(node * &x , int & k){
if(x->size == 0){
x = new node(k);
x->c[0] = nil;
x->c[1] = nil;
}else{
bool l = k > x->key;
x->size++;
insert(x->c[l] , k);
maintain(x , l);
};
};
inline void rot(node * & x , bool l){
node * y = x->c[!l];
x->c[!l] = y->c[l];
y->c[l] = x;
y->size = x->size;
x->t();
x = y;
};
inline void maintain(node * &x , bool l){
if(x->size == 0) return;
if(x->c[l]->c[l]->size > x->c[!l]->size){
rot(x , !l);
}else{
if(x->c[l]->c[!l]->size > x->c[!l]->size){
rot(x->c[l] , l);
rot(x , !l);
}else{
return;
};
};
maintain(x->c[1] , 1);
maintain(x->c[0] , 0);
maintain(x , 1);
maintain(x , 0);
};
int select(int n){
node* p=root;
while(n){
if(n<=p->c[0]->size){
p=p->c[0];
}else{
if(n==p->c[0]->size+1) break;
n-=p->c[0]->size+1;
p=p->c[1];
}
}
return p->key;
}
public:
SBT(){
nil = new node(0);
nil->size = 0;
nil->c[0] = nil;
nil->c[1] = nil;
root = nil;
};
inline void Insert(int k){
insert(root , k);
};
inline int Select(int n){
return select(n);
}
};
int a,i,j,b,k,c;
struct nod{
int a,id;
bool f;
} ned[600000];
bool cmp(nod a,nod b){
if(a.id<b.id){
return true;
}else{
if(a.id>b.id){
return false;
}else{
return a.f<b.f;
}
}
}
SBT sbt;
int pp;
int main(int argc, char const *argv[]){
freopen("blackbox.in","r",stdin);
freopen("blackbox.out","w",stdout);
scanf("%d%d",&a,&b);
for(i=0;i<a;++i){
scanf("%d",&ned[i].a);
ned[i].id=i+1;
ned[i].f=false;
}
for(b+=a,j=a;j<b;++j){
scanf("%d",&ned[j].id);
ned[j].f=true;
}
std::sort(ned,ned+j,cmp);
c=a;
a=0;
for(i=0;i<j;++i){
//printf("
%d : %d %d
",i,ned[i].f,ned[i].a);
if(ned[i].f){
printf("%d
",sbt.Select(++a));
}else{
sbt.Insert(ned[i].a);
++pp;
}
}
return 0;
}
这调试真爽啊!!tm花了我2天!!
还借鉴了Nocow上的代码!!
真乃 毒!品德芳香!大师!学习了! 烤焦了!太锐利了!犹如奶油一般绵延细长!在我的脑海里化开成为挥之不去的阴影!