题目链接:
UVALive 4043 Ants
题意:
给出平面上n个白点和n个黑点的点坐标,将这n个白点和n个黑点用n条不相交的线段连接,输出每个白点连接的黑点的编号.
n<=100.
分析:
假设有两个白点a1,a2和两个黑点b1,b2,且当前匹配为a1-b1,a2-b2.如果线段a1-b1和线段a2-b2相交,
可以得到距离关系dis(a1, b1) + dis(a2, b2) > dis(a1, b2) + dis(a2, b1).所以最佳的匹配是距离和短的匹配.
将白点和黑点建边,权值为距离的负数(求最小权匹配),然后用KM()求一下最大权匹配就好了.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#include <iomanip>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_N = 1010;
const double eps = 1e-8;
const double DOUBLE_MAX = 1e9;
const double DOUBLE_MIN = -1e9;int n, m, total;
int head[MAX_N], matchx[MAX_N], matchy[MAX_N], visx[MAX_N], visy[MAX_N];
double lx[MAX_N], ly[MAX_N], w[MAX_N][MAX_N], slack[MAX_N];
struct Point{double x, y;
}white[MAX_N], black[MAX_N];bool dfs(int x)
{visx[x] = 1;for(int y = 0; y < m; y++){if(visy[y]) continue;double tmp = lx[x] + ly[y] - w[x][y];if(fabs(tmp) <= eps){visy[y] = 1;if(matchy[y] == -1 || dfs(matchy[y])){matchx[x] = y;matchy[y] = x;return true;}}else {slack[y] = min(slack[y], tmp);}}return false;
}void KM()
{memset(matchy, -1, sizeof(matchy));memset(ly, 0, sizeof(ly));for(int i = 0; i < n; i++){lx[i] = DOUBLE_MIN;for(int j = 0; j < m; j++){lx[i] = max(lx[i], w[i][j]);}}for(int i = 0; i < n; i++){for(int j = 0; j < m; j++) { slack[j] = DOUBLE_MAX; }while(1){memset(visx, 0, sizeof(visx));memset(visy, 0, sizeof(visy));if(dfs(i)) break;double d = DOUBLE_MAX;for(int j = 0; j < m; j++){if(!visy[j]) d = min(d, slack[j]);}for(int j = 0; j < n; j++) { if(visx[j]) lx[j] -= d; }for(int j = 0; j < m; j++){if(visy[j]) ly[j] += d;else slack[j] -= d;}}}
}int main()
{IOS;int first = 1;while(cin >> n){if(first) first = 0;else cout << endl;m = n;for(int i = 0; i < n; i++){cin >> white[i].x >> white[i].y;}for(int i = 0; i < n; i++){cin >> black[i].x >> black[i].y;}for(int i = 0;i < n; i++){for(int j = 0; j < n; j++){w[i][j] = - hypot(white[i].x - black[j].x, white[i].y - black[j].y);//cout << fixed << setprecision(6) << w[i][j] << endl;}}KM();for(int i = 0; i < n; i++){cout << matchx[i] + 1 << endl;}}return 0;
}