Codeforces 196 C. Paint Tree
2016-01-19 20:04
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分治。选最左上的点分给根。剩下的极角排序后递归
C. Paint Tree
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a tree with n vertexes and n points
on a plane, no three points lie on one straight line.
Your task is to paint the given tree on a plane, using the given points as vertexes.
That is, you should correspond each vertex of the tree to exactly one point and each point should correspond to a vertex. If two vertexes of the tree are connected by an edge, then the corresponding points should have a segment painted between them. The segments
that correspond to non-adjacent edges, should not have common points. The segments that correspond to adjacent edges should have exactly one common point.
Input
The first line contains an integer n (1 ≤ n ≤ 1500)
— the number of vertexes on a tree (as well as the number of chosen points on the plane).
Each of the next n - 1 lines contains two space-separated integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi)
— the numbers of tree vertexes connected by the i-th edge.
Each of the next n lines contain two space-separated integers xi and yi ( - 109 ≤ xi, yi ≤ 109)
— the coordinates of thei-th point on the plane. No three points lie on one straight line.
It is guaranteed that under given constraints problem has a solution.
Output
Print n distinct space-separated integers from 1 to n:
the i-th number must equal the number of the vertex to place at the i-th
point (the points are numbered in the order, in which they are listed in the input).
If there are several solutions, print any of them.
Sample test(s)
input
output
input
output
Note
The possible solutions for the sample are given below.
C. Paint Tree
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given a tree with n vertexes and n points
on a plane, no three points lie on one straight line.
Your task is to paint the given tree on a plane, using the given points as vertexes.
That is, you should correspond each vertex of the tree to exactly one point and each point should correspond to a vertex. If two vertexes of the tree are connected by an edge, then the corresponding points should have a segment painted between them. The segments
that correspond to non-adjacent edges, should not have common points. The segments that correspond to adjacent edges should have exactly one common point.
Input
The first line contains an integer n (1 ≤ n ≤ 1500)
— the number of vertexes on a tree (as well as the number of chosen points on the plane).
Each of the next n - 1 lines contains two space-separated integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi)
— the numbers of tree vertexes connected by the i-th edge.
Each of the next n lines contain two space-separated integers xi and yi ( - 109 ≤ xi, yi ≤ 109)
— the coordinates of thei-th point on the plane. No three points lie on one straight line.
It is guaranteed that under given constraints problem has a solution.
Output
Print n distinct space-separated integers from 1 to n:
the i-th number must equal the number of the vertex to place at the i-th
point (the points are numbered in the order, in which they are listed in the input).
If there are several solutions, print any of them.
Sample test(s)
input
3 1 3 2 3 0 0 1 1 2 0
output
1 3 2
input
4 1 2 2 3 1 4 -1 -2 3 5 -3 3 2 0
output
4 2 1 3
Note
The possible solutions for the sample are given below.
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <vector> using namespace std; const int maxn=1520; int n,X,Y; struct PO { int x,y,d; bool operator<(const PO &o) const { if(x-X>=0&&o.x-X<=0) return 1; if(x-X<=0&&o.x-X>=0) return 0; return (y-Y)*(long long)(o.x-X)<(o.y-Y)*(long long)(x-X); } }p[maxn]; vector<int> g[maxn]; bool vis[maxn]; int sz[maxn],o[maxn]; int dfs(int u) { vis[u]=true; sz[u]=1; int ret=0; for(int i=0,j=g[u].size();i<j;i++) { int v=g[u][i]; if(vis[v]) continue; ret+=dfs(v); } sz[u]+=ret; return sz[u]; } void calc(int u,int l,int r) { vis[u]=true; int t=l; for(int i=l+1;i<=r;i++) { if((p[i].y<p[t].y)||(p[t].y==p[i].y&&p[i].x<p[t].x)) t=i; } if(t!=l) swap(p[l],p[t]); o[p[l].d]=u; X=p[l].x; Y=p[l].y; sort(p+l+1,p+r+1); int pos=l+1; for(int i=0,j=g[u].size();i<j;i++) { int v=g[u][i]; if(vis[v]) continue; calc(v,pos,pos+sz[v]-1); pos+=sz[v]; } } int main() { scanf("%d",&n); for(int i=0;i<n-1;i++) { int x,y; scanf("%d%d",&x,&y); g[x].push_back(y); g[y].push_back(x); } for(int i=1;i<=n;i++) { int x,y; scanf("%d%d",&x,&y); p[i]=(PO){x,y,i}; } dfs(1); memset(vis,0,sizeof(vis)); calc(1,1,n); for(int i=1;i<=n;i++) printf("%d ",o[i]); putchar(10); return 0; }
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