# ACM学习历程—POJ3565 Ants（最佳匹配KM算法）

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`

```#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <string>
#define LL long long

using namespace std;

const int maxN = 105;
const double inf = 1e10;
int n;
struct Point
{
double x, y;
}x[maxN], y[maxN];
int nx, ny;
bool visx[maxN], visy[maxN];

inline double dis(int i, int j)
{
double ans = (x[i].x-y[j].x)*(x[i].x-y[j].x) + (x[i].y-y[j].y)*(x[i].y-y[j].y);
return -sqrt(ans);
}

bool DFS(int x)
{
visx[x] = true;
for (int y = 1; y <= ny; y++)
{
if (visy[y])
continue;
double t = lx[x]+ly[y]-w[x][y];
if (fabs(t) < 1e-5)
{
visy[y] = true;
{
return true;
}
}
else if (slack[y] > t)//不在相等子图中slack取最小的
slack[y] = t;
}
return false;
}

void KM()
{
memset(ly, 0, sizeof(ly));
for (int i = 1; i <= nx; i++)//lx初始化为与它关联边中最大的
for (int j = 1; j <= ny; j++)
if (j == 1 || w[i][j] > lx[i])
lx[i] = w[i][j];

for (int x = 1; x <= nx; x++)
{
for (int i = 1; i <= ny; i++)
slack[i] = inf;
for (;;)
{
memset(visx, false, sizeof(visx));
memset(visy, false, sizeof(visy));
if (DFS(x))//若成功（找到了增广轨），则该点增广完成，进入下一个点的增广
break;//若失败（没有找到增广轨），则需要改变一些点的标号，使得图中可行边的数量增加。
//方法为：将所有在增广轨中（就是在增广过程中遍历到）的X方点的标号全部减去一个常数d，
//所有在增广轨中的Y方点的标号全部加上一个常数d
double d = inf;
for (int i = 1; i <= ny; i++)
if (!visy[i] && d > slack[i])
d = slack[i];
for (int i = 1; i <= nx; i++)
if (visx[i])
lx[i] -= d;
for (int i = 1; i <= ny; i++)//修改顶标后，要把所有不在交错树中的Y顶点的slack值都减去d
if (visy[i])
ly[i] += d;
else
slack[i] -= d;
}
}

for (int i = 1; i <= n; ++i)
}

void input()
{
nx = ny = n;
for (int i = 1; i <= n; ++i)
scanf("%lf%lf", &y[i].x, &y[i].y);
for (int i = 1; i <= n; ++i)
scanf("%lf%lf", &x[i].x, &x[i].y);
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
w[i][j] = dis(i, j);
}

int main ()
{
//freopen("test.in", "r", stdin);
while (scanf ("%d", &n) != EOF)
{
input();
KM();
}
return 0;
}```

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