Codeforces Round #394 (Div. 2) E. Dasha and Puzzle 构造 坐标离散化
2017-02-11 23:00
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题意:给一颗树,问是否能够放到网格图里,保证只有横边和竖边,且边与边不相交。
解法:
开脑洞,先想能不能找到这样的树,树应该至少满足一个要求:
1.根节点能有4个儿子,非根节点最多只有3个儿子。
满足要求1,理论上就一定能找到这样的树,即可以另深度为d的边比深度为d+1的边长很多很多,这样使得各个方向的子图离的尽量远,不管子树的图形再怎么乱,子树和子树之间也不会碰在一起。
满足这个要求的极限条件就是下面这个不等式:
2^0+2^1+...+2^(k-1)=2^k-1<2^k,可以刚好不碰到,故深度1-k的边长依次为2^k,2^(k-1),...,2^1,2^0。
构造完树之后,坐标离散化一下可以让节点坐标缩在0~n-1之间。
代码依次为去重离散化,未去重离散化,未离散化版本,题目给的范围不需要离散化也能过:
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned int uii;
const ll dx[4]={0,-1,0,1};
const ll dy[4]={-1,0,1,0};
int n,u,v,n1,n2;
ll rx[31],ry[31];
vector<ll> x,y;
vector<int> vec[31];
void dfs(int u,int fa,ll x,ll y,ll d,int dir) {
int id=0;
rx[u]=x;
ry[u]=y;
for (uii i=0;i<vec[u].size();++i) {
int v=vec[u][i];
if (v==fa)
continue;
if (dir!=-1&&((id&1)==(dir&1)&&id!=dir))
++id;
if (id>=4) {
puts("NO");
exit(0);
}
dfs(v,u,x+dx[id]*d,y+dy[id]*d,d>>1,id);
++id;
}
}
int main()
{
scanf("%d",&n);
for (int i=0;i<n-1;++i) {
scanf("%d%d",&u,&v);
vec[u].push_back(v);
vec[v].push_back(u);
}
dfs(1,-1,0,0,1LL<<n,-1);
for (int i=1;i<=n;++i) {
x.push_back(rx[i]);
y.push_back(ry[i]);
}
sort(x.begin(),x.end());
x.erase(unique(x.begin(),x.end()),x.end());
sort(y.begin(),y.end());
y.erase(unique(y.begin(),y.end()),y.end());
puts("YES");
for (int i=1;i<=n;++i)
printf("%ld %ld\n",lower_bound(x.begin(),x.end(),rx[i])-x.begin(),lower_bound(y.begin(),y.end(),ry[i])-y.begin());
return 0;
}
解法:
开脑洞,先想能不能找到这样的树,树应该至少满足一个要求:
1.根节点能有4个儿子,非根节点最多只有3个儿子。
满足要求1,理论上就一定能找到这样的树,即可以另深度为d的边比深度为d+1的边长很多很多,这样使得各个方向的子图离的尽量远,不管子树的图形再怎么乱,子树和子树之间也不会碰在一起。
满足这个要求的极限条件就是下面这个不等式:
2^0+2^1+...+2^(k-1)=2^k-1<2^k,可以刚好不碰到,故深度1-k的边长依次为2^k,2^(k-1),...,2^1,2^0。
构造完树之后,坐标离散化一下可以让节点坐标缩在0~n-1之间。
代码依次为去重离散化,未去重离散化,未离散化版本,题目给的范围不需要离散化也能过:
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned int uii;
const ll dx[4]={0,-1,0,1};
const ll dy[4]={-1,0,1,0};
int n,u,v,n1,n2;
ll rx[31],ry[31];
vector<ll> x,y;
vector<int> vec[31];
void dfs(int u,int fa,ll x,ll y,ll d,int dir) {
int id=0;
rx[u]=x;
ry[u]=y;
for (uii i=0;i<vec[u].size();++i) {
int v=vec[u][i];
if (v==fa)
continue;
if (dir!=-1&&((id&1)==(dir&1)&&id!=dir))
++id;
if (id>=4) {
puts("NO");
exit(0);
}
dfs(v,u,x+dx[id]*d,y+dy[id]*d,d>>1,id);
++id;
}
}
int main()
{
scanf("%d",&n);
for (int i=0;i<n-1;++i) {
scanf("%d%d",&u,&v);
vec[u].push_back(v);
vec[v].push_back(u);
}
dfs(1,-1,0,0,1LL<<n,-1);
for (int i=1;i<=n;++i) {
x.push_back(rx[i]);
y.push_back(ry[i]);
}
sort(x.begin(),x.end());
x.erase(unique(x.begin(),x.end()),x.end());
sort(y.begin(),y.end());
y.erase(unique(y.begin(),y.end()),y.end());
puts("YES");
for (int i=1;i<=n;++i)
printf("%ld %ld\n",lower_bound(x.begin(),x.end(),rx[i])-x.begin(),lower_bound(y.begin(),y.end(),ry[i])-y.begin());
return 0;
}
#include <cstdio> #include <vector> #include <cstdlib> #include <algorithm> using namespace std; typedef lo 4000 ng long ll; typedef unsigned int uii; const ll dx[4]={0,-1,0,1}; const ll dy[4]={-1,0,1,0}; int n,u,v,id1[31],id2[31]; ll rx[31],ry[31]; vector<int> vec[31]; bool cmp1(int a,int b) { return rx[a]<rx[b]; } bool cmp2(int a,int b) { return ry[a]<ry[b]; } void dfs(int u,int fa,ll x,ll y,ll d,int dir) { int id=0; rx[u]=x; ry[u]=y; for (uii i=0;i<vec[u].size();++i) { int v=vec[u][i]; if (v==fa) continue; if (dir!=-1&&((id&1)==(dir&1)&&id!=dir)) ++id; if (id>=4) { puts("NO"); exit(0); } dfs(v,u,x+dx[id]*d,y+dy[id]*d,d>>1,id); ++id; } } int main() { scanf("%d",&n); for (int i=1;i<=n;++i) id1[i]=id2[i]=i; for (int i=0;i<n-1;++i) { scanf("%d%d",&u,&v); vec[u].push_back(v); vec[v].push_back(u); } dfs(1,-1,0,0,1LL<<n,-1); sort(id1+1,id1+n+1,cmp1); sort(id2+1,id2+n+1,cmp2); puts("YES"); for (int i=1;i<=n;++i) printf("%ld %ld\n",lower_bound(id1+1,id1+n+1,i,cmp1)-id1,lower_bound(id2+1,id2+n+1,i,cmp2)-id2); return 0; }
#include <cstdio> #include <vector> #include <cstdlib> using namespace std; typedef long long ll; typedef unsigned int uii; const ll dx[4]={0,-1,0,1}; const ll dy[4]={-1,0,1,0}; int n,u,v; ll rx[31],ry[31]; vector<int> vec[31]; void dfs(int u,int fa,ll x,ll y,ll d,int dir) { int id=0; rx[u]=x; ry[u]=y; for (uii i=0;i<vec[u].size();++i) { int v=vec[u][i]; if (v==fa) continue; if (dir!=-1&&((id&1)==(dir&1)&&id!=dir)) ++id; if (id>=4) { puts("NO"); exit(0); } dfs(v,u,x+dx[id]*d,y+dy[id]*d,d>>1,id); ++id; } } int main() { scanf("%d",&n); for (int i=0;i<n-1;++i) { scanf("%d%d",&u,&v); vec[u].push_back(v); vec[v].push_back(u); } dfs(1,-1,0,0,1LL<<n,-1); puts("YES"); for (int i=1;i<=n;++i) printf("%lld %lld\n",rx[i],ry[i]); return 0; }
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