UVa 1411 Ants（分治）

https://vjudge.net/problem/UVA-1411

题意：n只蚂蚁和n颗苹果树，一一配对并且不能交叉。

思路：这就是巨人与鬼的问题。用分治法就行了。

#include<iostream>  
#include<algorithm>
#include<set>
using namespace std;

int n;
const int maxn = 205;

int vis[maxn];

struct node
{
    int x, y;
    int id;
    int flag;
}ans[maxn];


node s;

bool cmp1(node a, node b)   //按y坐标从小到大排序
{
    return a.y < b.y || (a.y == b.y && a.x < b.x);
}

bool cmp2(node a, node b)  //极角从小到大排序
{
    return ((a.x - s.x)*(b.y - s.y) - (a.y - s.y)*(b.x - s.x))<0;
}

void solve(int l,int r)
{
    if (l>r)  return;
    sort(ans + l, ans + r + 1, cmp1);
    s = ans[l];
    sort(ans + l + 1, ans + r + 1, cmp2);
    int cnt1 = 0, cnt2 = 0;
    int k = r;
    while (!(s.flag != ans[k].flag && cnt1 == cnt2))
    {
        if (ans[k].flag == s.flag)  cnt1++;
        else cnt2++;
        k--;
    }
    if (!s.flag)  vis[s.id] = ans[k].id;
    else  vis[ans[k].id] = s.id;
    solve(l + 1, k - 1);
    solve(k + 1, r);
}

int main()
{
    //freopen("D:\txt.txt", "r", stdin);
    while (cin >> n && n)
    {
        for (int i = 1; i <= n; i++)
        {
            cin >> ans[i].x >> ans[i].y;
            ans[i].id = i;   //蚂蚁编号
            ans[i].flag = 0;  //0代表蚂蚁
        }
        for (int i = n + 1; i <= 2 * n; i++)
        {
            cin >> ans[i].x >> ans[i].y;
            ans[i].id = i - n;  //苹果树编号
            ans[i].flag = 1;     //1代表苹果树
        }
        solve(1,2*n);
        for (int i = 1; i <= n; i++)
            cout << vis[i] << endl;
    }
    return 0;
}