Gym 100531H Problem H. Hiking in the Hills 二分
Problem H. Hiking in the Hills
题目连接：
http://codeforces.com/gym/100531/attachments
Description
Helen is hiking with her friends in a highland. Their plan is to hike from their camp A to a beautiful
showplace B.
Unfortunately, Helen started feeling dizzy due to altitude sickness. Help her group find a route such that
the topmost height on that route is as small as possible.
Input
The input file contains full information about the landscape of a square region 106 × 106
in the following
format. The first line contains integer n — the number of triangles in the landscape (2 ≤ n ≤ 2000).
Each of following n lines contains nine integers xi1, yi1, zi1, xi2, yi2, zi2, xi3, yi3, zi3 — coordinates of a
triangle. All coordinates belong to the closed interval [0, 106
]. The two last lines contain three integers
each: xA, yA, zA and xB, yB, zB — coordinates of the camp A and the showplace B.
The given triangles are guaranteed to describe a consistent continuous landscape. Projections of triangles
onto XY plane are non-degenerate and fill the square without overlapping. A vertex of one triangle never
lays inside an edge of another triangle. Points A and B belong to the landscape surface and are different.
Output
Output a polyline route from A to B with the smallest possible topmost height. The first line should
contain m, the number of vertices in this polyline. Each of following m lines should contain three integer
coordinates of a polyline vertex: xi
, yi
, and zi
. Vertices must be listed along the polyline, from A to B
(including these two endpoints).
All coordinates of polyline vertices should be integer. Each polyline edge must belong to some triangle
from the input file (possibly, to its edge). The number of vertices in the polyline must not exceed 5n.
Sample Input
8
1000000 0 0 1000000 1000000 150000 600000 600000 400000
0 1000000 0 600000 600000 400000 600000 1000000 300000
0 1000000 0 400000 300000 150000 600000 600000 400000
400000 0 200000 1000000 0 0 400000 300000 150000
400000 300000 150000 1000000 0 0 600000 600000 400000
600000 600000 400000 1000000 1000000 150000 600000 1000000 300000
0 0 0 400000 0 200000 400000 300000 150000
0 1000000 0 0 0 0 400000 300000 150000
100000 700000 37500
900000 400000 137500
Sample Output
4
100000 700000 37500
400000 300000 150000
900000 150000 100000
900000 400000 137500
Hint
题意
给你一个多面体，每个平面都是一个三角形
然后给你一个A点和B点，你需要输出一个从A到B的路径，使得这条路径的最高点最低
题解：
首先，走点一定是可行的，所以我们就可以不用去考虑边。
在一个三角形内的话，就连一条边。
然后我们直接二分高度，然后每次CHECK A是否能到B 就好了
注意精度有毒。。。
代码
#include<bits/stdc++.h>
using namespace std;
struct node
{
double x,y,z;
bool operator<(const node& p) const
{
if(z==p.z&&y==p.y)return x<p.x;
if(z==p.z)return y<p.y;
return z<p.z;
}
};
struct Tri
{
node p[3];
};
Tri tri[5006];
map<node,int> H;
map<int,node> T;
int tot = 1;
node A,B;
vector<int> E[7000];
int vis[7000];
int n;
void init()
{
memset(vis,0,sizeof(vis));
H.clear();
T.clear();
tot = 1;
for(int i=0;i<7000;i++)
E[i].clear();
memset(tri,0,sizeof(tri));
}
double eps = 1e-2;
double dis(node aa,node bb)
{
return sqrt((aa.x-bb.x)*(aa.x-bb.x)+(aa.y-bb.y)*(aa.y-bb.y)+(aa.z-bb.z)*(aa.z-bb.z));
}
double area(node aa,node bb,node cc)
{
double l1 = dis(aa,bb);
double l2 = dis(aa,cc);
double l3 = dis(bb,cc);
double pp = (l1+l2+l3)/2.0;
return sqrt(pp*(pp-l1)*(pp-l2)*(pp-l3));
}
int inRan(Tri kkk,node ttt)
{
double a1 = area(kkk.p[0],kkk.p[1],ttt);
double a2 = area(kkk.p[0],kkk.p[2],ttt);
double a3 = area(kkk.p[1],kkk.p[2],ttt);
double a4 = area(kkk.p[0],kkk.p[1],kkk.p[2]);
if(fabs(a4-a1-a2-a3)<=eps)return 1;
return 0;
}
void dfs(int x,int h)
{
vis[x]=1;
for(int i=0;i<E[x].size();i++)
{
int v = E[x][i];
if(vis[v])continue;
if(T[v].z>h)continue;
dfs(v,h);
}
}
int check(double h)
{
if(A.z>h||B.z>h)return 0;
memset(vis,0,sizeof(vis));
dfs(H[A],h);
if(vis[H[B]]==1)return 1;
return 0;
}
vector<node> TTT;
int flag = 0;
void dfs2(int x,double h)
{
if(flag)return;
TTT.push_back(T[x]);
if(x==H[B])
{
flag = 1;
cout<<TTT.size()<<endl;
for(int i=0;i<TTT.size();i++)
printf("%.0f %.0f %.0f\n",TTT[i].x,TTT[i].y,TTT[i].z);
return;
}
vis[x]=1;
for(int i=0;i<E[x].size();i++)
{
int v = E[x][i];
if(vis[v])continue;
if(T[v].z>h)continue;
dfs2(v,h);
TTT.pop_back();
}
}
int main()
{
freopen("hiking.in","r",stdin);
freopen("hiking.out","w",stdout);
init();
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
for(int j=0;j<3;j++)
{
scanf("%lf%lf%lf",&tri[i].p[j].x,&tri[i].p[j].y,&tri[i].p[j].z);
if(H[tri[i].p[j]]==0)
{
T[tot] = tri[i].p[j];
H[tri[i].p[j]] = tot++;
}
}
for(int j=0;j<3;j++)
{
for(int k=j+1;k<3;k++)
{
E[H[tri[i].p[j]]].push_back(H[tri[i].p[k]]);
E[H[tri[i].p[k]]].push_back(H[tri[i].p[j]]);
}
}
}
scanf("%lf%lf%lf",&A.x,&A.y,&A.z);
scanf("%lf%lf%lf",&B.x,&B.y,&B.z);
if(H[A]==0)
{
T[tot] = A;
H[A] = tot++;
for(int i=1;i<=n;i++)
{
if(inRan(tri[i],A))
{
for(int j=0;j<3;j++)
{
E[H[A]].push_back(H[tri[i].p[j]]);
E[H[tri[i].p[j]]].push_back(H[A]);
}
}
}
}
if(H[B]==0)
{
T[tot] = B;
H[B] = tot++;
for(int i=1;i<=n;i++)
{
if(inRan(tri[i],B))
{
for(int j=0;j<3;j++)
{
E[H[B]].push_back(H[tri[i].p[j]]);
E[H[tri[i].p[j]]].push_back(H[B]);
}
}
}
}
double l = -2.0,r = 3000050.0;
for(int i=1;i<=100;i++)
{
double mid = (l+r)/2.0;
if(check(mid))r=mid;
else l=mid;
}
memset(vis,0,sizeof(vis));
TTT.clear();
dfs2(H[A],r+1);
}
