POJ： 2236 Wireless Network 题解

POJ 2236 Wireless Network

加工并储存数据的数据结构

并查集

这是并查集的基本应用，两台修好的电脑若距离d内则加入合并。不过不小心的话会TLE，比如：

#include <iostream>
using namespace std;
 
#define MAX_N 1001 + 16
int parent[MAX_N];
int height[MAX_N];
bool status[MAX_N];
int distance[MAX_N][MAX_N];
 
void init(const int& n)
{
	for (int i = 0; i < n; ++i)
	{
		parent[i] = i;
		height[i] = 0;
	}
}
 
int find(const int& x)
{
	if (parent[x] == x)
	{
		return x;
	}
	else
	{
		return parent[x] = find(parent[x]);
	}
}
 
void unite(int x, int y)
{
	x = find(x);
	y = find(y);
	if (x == y)
	{
		return;
	}
 
	if (height[x] < height[y])
	{
		parent[x] = y;
	}
	else
	{
		parent[y] = x;
		if (height[x] == height[y])
		{
			++height[x];
		}
	}
}
 
bool same(const int& x, const int& y)
{
	return find(x) == find(y);
}
 
pair<int, int> computer[MAX_N];
int square(const int& x)
{
	return x * x;
}

int main(int argc, char *argv[])
{
	int N, d;
	cin >> N >> d;
	for (int i = 0; i < N; ++i)
	{
		cin >> computer[i].first >> computer[i].second;
	}
	init(N);
	char operation;
	int x, y;
	while (cin >> operation)
	{
		if (operation == 'O')
		{
			cin >> x;
			--x;
			status[x] = true;
			for (int i = 0; i < N; ++i)
			{
				if (i == x)
				{
					continue;
				}
				if (status[i] && square(computer[x].first - computer[i].first) + square(computer[x].second - computer[i].second) <= square(d))
				{
					unite(x, i);
				}
			}
		}
		else
		{
			cin >> x >> y;
			--x; --y;
			if (same(x, y))
			{
				cout << "SUCCESS" << endl;
			}
			else
			{
				cout << "FAIL" << endl;
			}
		}
	}
	return 0;
}

平方计算太多了，初始化的时候算一次记录在一个二维数组中就够了。

#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;

#define ms(a,b) memset(a,b,sizeof(a));
#define maxn  1010

int f[maxn];
int h[maxn];
bool status[maxn];//电脑是否维修好了
bool able[maxn][maxn];//distance

void init(int n) {
    //电脑编号从0开始
    for (int i = 0; i < n; ++i) {
        f[i] = i;
        h[i] = 0;
    }
}

int find(int x) {
    return f[x] == x ? x : f[x] = find(f[x]);
}

void merge(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y)return;
    if (h[x] < h[y])f[x] = y;
    else {
        f[y] = x;
        if (h[x] == h[y])++h[x];
    }
}

bool same(int x, int y) {
    return find(x) == find(y);
}

pair<int, int>computer[maxn];
int square(int x) {
    return x * x;
}

int main() {
    int N, d;
    cin >> N >> d;
    for (int i = 0; i < N; ++i)
    {
        cin >> computer[i].first >> computer[i].second;
    }
    init(N);
    for (int i = 0; i < N; ++i)
    {
        for (int x = i; x < N; ++x)
        {
            if (square(computer[x].first - computer[i].first) + square(computer[x].second - computer[i].second) <= square(d))
            {
                able[i][x] = true;
                able[x][i] = true;
            }
        }
    }
    char operation;
    int x, y;
    while (cin >> operation) {
        if (operation == 'O')
        {
            cin >> x;
            --x;
            status[x] = true;
            for (int i = 0; i < N; ++i)
            {
                if (i == x)
                {
                    continue;
                }
                if (status[i] && able[x][i])
                {
                    merge(x, i);
                }
            }
        }
        else
        {
            cin >> x >> y;
            --x; --y;
            if (same(x, y))
            {
                cout << "SUCCESS" << endl;
            }
            else
            {
                cout << "FAIL" << endl;
            }
        }
    }
}

标签：return,int,题解,cin,Wireless,++,computer,POJ,find
来源： https://www.cnblogs.com/RioTian/p/12864935.html