| Time Limit: 5000MS | Memory Limit: 65536K | |||
| Total Submissions: 4251 | Accepted: 1309 | Special Judge | ||
Description
Young naturalist Bill studies ants in school. His ants feed on plant-louses that live on apple trees. Each ant colony needs its own apple tree to feed itself.
Bill has a map with coordinates of n ant colonies and n apple trees. He knows that ants travel from their colony to their feeding places and back using chemically tagged routes. The routes cannot intersect each other or ants will get confused and get to the wrong colony or tree, thus spurring a war between colonies.
Bill would like to connect each ant colony to a single apple tree so that all n routes are non-intersecting straight lines. In this problem such connection is always possible. Your task is to write a program that finds such connection.
On this picture ant colonies are denoted by empty circles and apple trees are denoted by filled circles. One possible connection is denoted by lines.
Input
The first line of the input file contains a single integer number n (1 ≤ n ≤ 100) — the number of ant colonies and apple trees. It is followed by n lines describing n ant colonies, followed by n lines describing n apple trees. Each ant colony and apple tree is described by a pair of integer coordinates x and y (−10 000 ≤ x, y ≤ 10 000) on a Cartesian plane. All ant colonies and apple trees occupy distinct points on a plane. No three points are on the same line.
Output
Write to the output file n lines with one integer number on each line. The number written on i-th line denotes the number (from 1 to n) of the apple tree that is connected to the i-th ant colony.
Sample Input
5 -42 58 44 86 7 28 99 34 -13 -59 -47 -44 86 74 68 -75 -68 60 99 -60
Sample Output
4 2 1 5 3
Source
/*
題意：n個螞蟻和n棵樹,給出他們的座標，現在n個螞蟻走到這n個樹下產生n條路徑
問你求出一種方案使之螞蟻們的路徑之間不能交叉（給出一種合理的方案即可）
此題開始我們可以看成螞蟻和樹的配對是亂序的（假設螞蟻i配對上樹j）然後我們不斷匹配不斷檢查當前狀態是否滿足題目要求
若發現有配對線是交叉的，那麼我們就對其進行交換，使其不再交叉，到最後一定可以找出不交叉的路徑
*/
#include<iostream>
#include<queue>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iomanip>
#include<map>
#include<cstdlib>
#include<cmath>
#include<vector>
#define LL long long
#define IT __int64
#define zero(x) fabs(x)<eps
#define mm(a,b) memset(a,b,sizeof(a))
const int INF=0x7fffffff;
const double inf=1e8;
const double eps=1e-10;
const double PI=acos(-1.0);
const int Max=210;
using namespace std;
int sign(double x)
{
return (x>eps)-(x<-eps);
}
typedef struct Node
{
int x;
int y;
Node(const int &_x=0, const int &_y=0) : x(_x), y(_y) {}
void input()
{
cin>>x>>y;
}
void output()
{
cout<<x<<" "<<y<<endl;
}
}point;
point ant[Max],tree[Max];
int res[Max];
double xmult(point p0,point p1,point p2)
{
return(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
bool IsIntersected(point s1,point e1,point s2,point e2)
{
return (max(s1.x,e1.x)>=min(s2.x,e2.x))&& (max(s2.x,e2.x)>=min(s1.x,e1.x))&&(max(s1.y,e1.y)>=min(s2.y,e2.y))&&(max(s2.y,e2.y)>=min(s1.y,e1.y))&&(xmult(s1,s2,e1)*xmult(s1,e1,e2)>=0)&&(xmult(s2,s1,e2)*xmult(s2,e2,e1)>=0);
}
int main()
{
int n,m,i,j;
while(cin>>n)
{
for(i=1;i<=n;i++)
ant[i].input();
for(j=1;j<=n;j++)
tree[j].input();
for(i=1;i<=n;i++)
res[i]=i;
while(true)
{
int ok=1;
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
{
if(i==j)
continue;
if(IsIntersected(ant[i],tree[res[i]],ant[j],tree[res[j]]))
{//不斷檢查當前狀態是否滿足題目要求，若發現有配對線是交叉的，那麼我們就對其進行交換，使其不再交叉
swap(res[i],res[j]);
ok=0;
}
}
}
if(ok)//如果找出了沒有相交線路直接跳出
break;
}
for(i=1;i<=n;i++)
cout<<res[i]<<endl;
}
return 0;
}
