题意：在坐标系中有N只蚂蚁，N棵苹果树，给你蚂蚁和苹果树的坐标。让每只蚂蚁去一棵苹果树，一棵苹果树对应一只蚂蚁。这样就有N条直线路线，问：怎样分配，才能使总路程和最小，且N条线不相交。
分析：不相交线段长度（取反）转化为最大匹配，详见《算法竞赛进阶指南》P427。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int N = 105;
int a[N], b[N], c[N], d[N];
double w[N][N]; // 边权
double la[N], lb[N]; // 左、右部点的顶标
bool va[N], vb[N]; // 访问标记：是否在交错树中
int match[N], ans[N]; // 右部点匹配了哪一个左部点
int n;
double upd[N], delta;bool dfs(int x) {va[x] = 1; // 访问标记：x在交错树中for (int y = 1; y <= n; y++)if (!vb[y])if (fabs(la[x] + lb[y] - w[x][y]) < 1e-8) { // 相等子图vb[y] = 1; // 访问标记：y在交错树中if (!match[y] || dfs(match[y])) {match[y] = x;return true;}}else upd[y] = min(upd[y], la[x] + lb[y] - w[x][y]);return false;
}void KM() {for (int i = 1; i <= n; i++) {la[i] = -1e10; // -inflb[i] = 0;for (int j = 1; j <= n; j++)la[i] = max(la[i], w[i][j]);}for (int i = 1; i <= n; i++)while (true) { // 直到左部点找到匹配memset(va, 0, sizeof(va));memset(vb, 0, sizeof(vb));delta = 1e10; // inffor (int j = 1; j <= n; j++) upd[j] = 1e10; if (dfs(i)) break;for (int j = 1; j <= n; j++)if (!vb[j]) delta = min(delta, upd[j]);for (int j = 1; j <= n; j++) { // 修改顶标if (va[j]) la[j] -= delta;if (vb[j]) lb[j] += delta;}}
}int main()
{cin >> n;for (int i = 1; i <= n; i++) scanf("%d%d", &a[i], &b[i]);for (int i = 1; i <= n; i++) scanf("%d%d", &c[i], &d[i]);for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)w[i][j] = -sqrt((a[i]-c[j])*(a[i]-c[j])*1.0+(b[i]-d[j])*(b[i]-d[j])*1.0);KM();for (int i = 1; i <= n; i++) ans[match[i]] = i;for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);return 0;
}