## poj 2565 Ants (KM+思维)

2019年04月13日 算法 ⁄ 共 3187字 ⁄ 字号 评论关闭
Ants
 Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 4125 Accepted: 1258 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 ≤ xy ≤ 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

```#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-6;
const double PI=acos(-1.0);
const int maxn=110;
//typedef __int64 ll;
int le[maxn],n;
double lx[maxn],ly[maxn],slack[maxn],w[maxn][maxn],ax[maxn],ay[maxn],bx[maxn],by[maxn];
bool visx[maxn],visy[maxn];
double dist(double x1,double y1,double x2,double y2)
{
double x=x1-x2,y=y1-y2;
return sqrt(x*x+y*y);
}
bool match(int x)
{
int y;
double tp;
visx[x]=true;
for(y=1;y<=n;y++)
{
if(visy[y])
continue;
tp=lx[x]+ly[y]-w[x][y];
if(tp<eps)
{
visy[y]=true;
if(!le[y]||match(le[y]))
{
le[y]=x;
return true;
}
}
else
slack[y]=min(slack[y],tp);
}
return false;
}
void update()
{
int i;
double d;

for(i=1,d=INF;i<=n;i++)
if(!visy[i])
d=min(d,slack[i]);
for(i=1;i<=n;i++)
{
if(visx[i])
lx[i]-=d;
if(visy[i])
ly[i]+=d;
else
slack[i]-=d;
}
}
void KM()
{
int i,j,x;

memset(le,0,sizeof le);
for(i=1;i<=n;i++)
{
lx[i]=-INF;//注意这里!!
ly[i]=0;
for(j=1;j<=n;j++)
lx[i]=max(lx[i],w[i][j]);
}
for(x=1;x<=n;x++)
{
for(i=1;i<=n;i++)
slack[i]=INF;
while(1)
{
for(i=1;i<=n;i++)
visx[i]=visy[i]=false;
if(match(x))
break;
else
update();
}
}
}
int main()
{
int i,j;

while(~scanf("%d",&n))
{
for(i=1;i<=n;i++)
scanf("%lf%lf",&ax[i],&ay[i]);
for(i=1;i<=n;i++)
scanf("%lf%lf",&bx[i],&by[i]);
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
w[j][i]=-dist(ax[i],ay[i],bx[j],by[j]);//这里特别注意!由于答案输出的关系
KM();
for(i=1;i<=n;i++)
printf("%d\n",le[i]);
}
return 0;
}
```