[题解] poj 3565 Ants (二分图最佳完美匹配)

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　http://poj.org/problem?id=3565

#Ants | **Time Limit:** 5000MS |   | **Memory Limit:** 65536K | | **Total Submissions:** 6758 |   | **Accepted:** 2131 |   | 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

[[Submit](http://poj.org/submit?problem_id=3565)]   [[Status](http://poj.org/problemstatus?problem_id=3565)]   [[Discuss](http://poj.org/bbs?problem_id=3565)] 　 ### - 题意 - 　给出 n 个白点, n 个黑点的坐标, 要用 n 条不相连的线段将它们连起来, 每条线段连一个白点和一个黑点. 　 ### - 思路 - 　终点在于我们要使答案最小, 因为如果对于白点(A1, A2)和黑点(B1, B2), A1-B1 与 A2-B2 相交的话, A1-B2 加 A2-B1 一定会更小, (可以画图证明.) 所以最小解一定是一个答案. 　理解之后就是裸的二分图匹配了. 　二分图最佳完美匹配求最大答案, 所以我们要对边权取反来操作. 　 　细节见代码. 　 ### - 代码 - ```c++ #include #include #include #include using namespace std;

typedef double db;
const int N = 105;
const double inf = 1000000000;
const double eps = 1e-6;

db MAP[N][N], cx[N], cy[N];
int line[N], usex[N], usey[N];
int X[N], Y[N];
int n;
int x, y;

db mk(int x1, int x2, int y1, int y2) {
return -1.0 * sqrt((x1 - x2)(x1 - x2) + (y1 - y2)(y1 - y2));
}

bool find(int x) {
usex[x] = 1;
for (int i = 1; i <= n; ++i) {
if (usey[i] == 0 && fabs(cx[x] + cy[i] - MAP[x][i]) <= eps) {
usey[i] = 1;
if (line[i] == 0 || find(line[i])) {
line[i] = x;
return true;
}
}
}
return false;
}

void km() {
for (int i = 1; i <= n; ++i) {
while (true) {
db d = inf;
memset(usex, 0, sizeof (usex));
memset(usey, 0, sizeof (usey));
if (find(i)) break;
for (int j = 1; j <= n; ++j) {
if (usex[j])
for (int k = 1; k <= n; ++k)
if (!usey[k] && d > cx[j] + cy[k] - MAP[j][k])
d = cx[j] + cy[k] - MAP[j][k];
}
if (d == inf) return;
for (int j = 1; j <= n; ++j)
if (usex[j]) cx[j] -= d;
for (int j = 1; j <= n; ++j)
if (usey[j]) cy[j] += d;
}
}
for (int i = 1; i <= n; ++i) {
printf("%d ", line[i]);
}
}

int main() {
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf("%d%d", &X[i], &Y[i]);
}
for (int i = 1, x, y; i <= n; ++i) {
scanf("%d%d", &x, &y);
db d = 0;
for (int j = 1; j <= n; ++j) {
MAP[i][j] = mk(X[j], x, Y[j], y);
if (MAP[i][j] > d) d = MAP[i][j];
}
cx[i] = d;
}
km();
return 0;
}

  [1]: http://www.cnblogs.com/Anding-16/p/7410725.html