Tactical Multiple Defense System 二分图

This problem is about a war game between two countries. To protect a base, your country built a defense system called “Tactical Multiple Defense System” (TMD system in short). There are two weapons in the TMD system: Line gun and Circle gun. The Line gun can in one shot destroy all targets whose (two-dimensional) coordinates are on the same ray from the base, and the Circle gun can in one shot destroy all the targets with the same distance to the base. Note that in this game the coordinate of the base is (0, 0).

The other country is going to attack the base of your country. They deploy missiles at some places according to their “National Missile Deployment Plan” (NMD plan). Your spy got the NMD plan and therefore you have the positions of all the missiles, which are the targets you need to destroy. As the commander of the TMD system, your mission is to determine how to destroy all the n missiles by Line gun and Circle gun with minimum number of total shots.

n and Circle gun with minimum number of total shots. The position Pi of a missile is given by three positive integers ri , si , ti which indicates the polar coordinate is (ri , arctan(ti/si)), i.e., the distance from the base to Pi is ri and the slope of the ray from the base and through Pi is ti/si . We shall say that Pi is on the ray of slope ti/si . To use the Line gun, you input two integer parameters t and s, press the fire button, and then it destroys all targets (missiles) on the ray of slope t/s. On the other hand, to use the Circle gun, you need to input a positive integer parameter r, and it can destroy all targets with distance r to the base, that is, it destroys targets exactly on the circle of radius r (but not the ones within the circle). Figure 8 illustrates some examples.

 

Technical Specification

•The number of missiles n is at most 20000 in each test case. It is possible that two missiles are at the same position.

• The three parameters (ri , si , ti) of each position are integers and satisfy 1000 < ri ≤ 6000 and 1 ≤ si , ti ≤ 10000.

 

Input

The first line contains an integer T indicating the number of test cases. There are at most 10 test cases. For each test case, the first line is the number of missiles n. Each of the next n lines contains the parameters ri , si , ti of one missile, and two consecutive integers are separated by a space.

Output

For each test case, output in one line the minimum number of shots to destroy all the missiles.

 

Sample Input

1

5

1010 1 2

1020 2 4

1030 3 6

1030 9 9

1030 9 1

 

Sample Output

2

 

思路：

给出n个点距离原点（0,0）的长度ri和在直角坐标系的横纵坐标，有两种方法课一下消灭这些点,1是从原点发出一条一定角度的射线，2是发出一条一定半径的弧线，在这些线上的点都会被消灭，问最少需要多少条线可以将全部点消灭

可以假设任何一个点都经过一条射线和一条弧线，那么所有的点都会成为射线和弧线的交点，题目则转化为需要最少多少条这些线可以将所有的这些点都覆盖，（以线为点，两种线的交点为边建图， 即是求最小嗲覆盖），可以将射线当做x集合，弧线当做y集合，给所有的射线和弧线分别用map标号，再跑一变二分图即可

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <vector>
#include <cstdlib>
using namespace std;
typedef long long ll;

const int N = 22222;
int mat[N], vis[N], pos1[N], pos2[N], cnt, n;
vector<int> G[N];

bool Crosspath(int k)
{
    int sx = G[k].size();
    for(int i = 0; i < sx; ++i)
    {
        int j = G[k][i];
        if(vis[j]) continue;
        vis[j] = 1;
        if(mat[j] == -1 || Crosspath(mat[j])) {
            mat[j] = k;
            return true;
        }
    }
    return false;
}

void hungary()
{
    cnt = 0;
    memset(mat, -1, sizeof mat);
    for(int i = 1; i < n; ++i)
    {
        memset(vis, 0, sizeof vis);
        if(Crosspath(i)) cnt++;
    }
    printf("%d\n", cnt);
}

int main()
{
    int _;
    scanf("%d", &_);
    while(_ --)
    {
        scanf("%d", &n);
        map<int, int> mp1;
        map<double, int> mp2;
        mp1.clear();
        mp2.clear();
        for(int i = 0; i <= n; ++i) G[i].clear();
        int num1 = 1, num2 = 1;
        int a, b, c;
        for(int i = 0; i < n; ++i) {
            scanf("%d%d%d", &a, &b, &c);
            double deg = c * 1.0 / b;
            if(mp2[deg])  pos2[i] = mp2[deg];
            else { pos2[i] = num2; mp2[deg] = num2++; }

            if(mp1[a]) pos1[i] = mp1[a];
            else { pos1[i] = num1; mp1[a] = num1++; }
        }

        for(int i = 0; i < n; ++i) {
            G[ pos1[i] ].push_back(pos2[i]);

        }
    //    cout << num1 << endl;
        n = num1;
    //    for(int i = 0; i < n; ++i) printf("%d ", pos1[i]);
    //    cout << endl;
    //    for(int i  = 0; i < n; ++i) printf("%d ", pos2[i]);
        hungary();
    }
}

 

　　

 

转载于:https://www.cnblogs.com/orchidzjl/p/4756469.html