2018 ACM-ICPC 南京
文章目录
- 2018 ACM-ICPC 南京站
- A Adrien and Austin
- D. Country Meow
- Problem G. Pyramid
- Problem I. Magic Potion
- Problem J. Prime Game
- Problem K. Kangaroo Puzzle
2018 ACM-ICPC 南京站
A Adrien and Austin
这是https://vjudge.net/contest/255624#status//M/0/ 的简化
if(K == 1)
printf("%s\n",(N&1)?"Adrien":"Austin");
else
printf("%s\n",(N!= 0)?"Adrien":"Austin");
D. Country Meow
三分算法枚举圆心的位置
const int maxn = 100+10;
struct Point{
double x,y,z;
};
Point p[maxn];
double a[maxn];
int n;
inline double mysqr(double x){
return x*x;
}
double cal(){
double ans = 0;
for(int i = 1;i <= n; ++i)
{
ans = max(ans,sqrt(mysqr(p[i].x-a[0])+mysqr(p[i].y-a[1])+mysqr(p[i].z-a[2])));
}
return ans;
}
double check(int cnt ){
if(cnt >= 3) return cal();
double l = -100000,r = 100000;
double ans = 1e10;
for(int i = 0;i < 50; ++i){
double mid1 = (2*l+r)/3;
double mid2 = (l+2*r)/3;
a[cnt] = mid1;
double ans1 = check(cnt+1);
a[cnt] = mid2;
double ans2 = check(cnt+1);
if(ans1 < ans2)
r = mid2,ans = min(ans,ans1);
else
l = mid1,ans = min(ans,ans2);
}
return ans;
}
int main(void)
{
cin>>n;
for(int i =1;i<= n; ++i)
scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].z);
printf("%.10f\n",check(0));
return 0;
}
Problem G. Pyramid
C(n+3,4) 不要问我为什么心痛推不出来
Problem I. Magic Potion
最大流
首先建源点s和汇点t
s - > s1 容量为n
s -> s2 容量为k
然后再从s1和s2 连到n个点，之后把该加的边加上
const double eps = 1e-6;
const LL mod = 1e9 + 7;
const int LEN = 20000+1000;
const int maxn = 1e8;
struct Edge{
int from,to,cap,flow;
Edge(int u,int v,int w,int f): from(u),to(v),cap(w),flow(f){}
};
struct Dinic{
int n,m,s,t;
vector<Edge> edges;
vector<int> G[LEN];
int a[LEN];
int vis[LEN];
int d[LEN];
int cur[LEN];//好吧就是点, 代表该点在一次求增广的过程中搜索到了那条边,意思就是从这条边往下肯定搜索不到结果了
void init(int n)
{
this->n = n;
for(int i = 0;i < n; ++i)
G[i].clear();
edges.clear();
}
void Add(int u,int v,int w)
{
edges.push_back(Edge(u,v,w,0));
edges.push_back(Edge(v,u,0,0));
m = edges.size();
G[u].push_back(m-2);
G[v].push_back(m-1);
}
bool Bfs(void)//分层
{
me(d);
me(vis);
d[s] = 0;
vis[s] = 1;
queue<int> Q;
Q.push(s);
while(!Q.empty())
{
int q = Q.front();Q.pop();
for(size_t i = 0;i < G[q].size();++i)
{
Edge &tmp = edges[G[q][i]];
if(!vis[tmp.to]&&tmp.cap>tmp.flow)
{
vis[tmp.to] = 1;
d[tmp.to] = d[q] + 1;
Q.push(tmp.to);
}
}
}
return vis[t];
}
int Dfs(int node,int a)
{
if(node == t||a == 0)
return a;
int flow = 0,f;
for(int &i = cur[node];i < (int)G[node].size();++i)
{
Edge &tmp = edges[G[node][i]];
if(d[tmp.to]==d[node]+1&&(f=Dfs(tmp.to,min(a,tmp.cap-tmp.flow)))>0)
{
flow += f;
tmp.flow += f;
edges[G[node][i]^1].flow -= f;
a -= f;
if(a==0)
break;
}
}
return flow;
}
int MaxFlow(int s,int t)
{
this->s = s;
this->t = t;
int flow = 0;
while(Bfs())
{
me(cur);
flow += Dfs(s,maxn);
}
return flow;
}
};
Dinic dinic;
int main()
{
int N,M,K;
scanf("%d%d%d",&N,&M,&K);
int d1 = N+M+1;
int d2 = N+M+2;
int dest = N+M+3;
int d3 = N+M+4;
dinic.init(N+M+10);
for(int i = 1;i <= N; ++i)
{
int t;
scanf("%d",&t);
for(int j = 1;j <= t; ++j){
int v;
scanf("%d",&v);
dinic.Add(i,v+N,1);
}
}
for(int i = 1;i <= N; ++i)
dinic.Add(d1,i,1),dinic.Add(d2,i,1);
dinic.Add(d3,d1,N);
dinic.Add(d3,d2,K);
for(int i = 1;i <= M; ++i)
dinic.Add(N+i,dest,1);
int ans = dinic.MaxFlow(d3,dest);
printf("%d\n",ans);
return 0;
}
Problem J. Prime Game
记录下这个素数出现的位置就行了
#include
for(int i = 1;i <= n; ++i){
// cout<<"DEBUG"<
for(int j = 1;Prime[j]*Prime[j] <= a[i] &&j < tot; ++j){
if(a[i] % Prime[j] == 0){
vec[Prime[j]].Pb(i);
while(a[i] % Prime[j] == 0) a[i] /= Prime[j];
}
}
if(a[i] != 1)
vec[a[i]].Pb(i);
}
LL ans = 0;
for(int i = 1;i < maxn; ++i){
if(vec[i].empty()) continue;
int l = 0;
for(int j = 0;j < (int) vec[i].size(); ++j)
{
ans += 1ll*(vec[i][j]-l)*(n+1-vec[i][j]);
l = vec[i][j];
}
}
cout<<ans<<endl;
return 0;
}
Problem K. Kangaroo Puzzle
int main(void)
{
int n;
srand(time(0));
n = 50000;
string s;
char a[5]= "UDRL";
while(n--) s += a[rand()%4];
cout<<s<<endl;
return 0;
}