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

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

int f[maxn];
int h[maxn];
pair<int, int>dis[maxn];
bool status[maxn];//电脑是否维修好了
bool able[maxn][maxn];//distance

void init() {
    for (int i = 0; i < maxn; ++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 a, int b) {
    return find(a) == find(b);
}

int square(int x) {
    return x * x;
}

int main() {
    int N, d;
    cin >> N >> d;
    for (int i = 0; i < N; ++i)cin >> dis[i].first >> dis[i].second;
    init();
    for (int i = 0; i < N; ++i)for (int x = i; x < N; ++x)
        if (square(dis[x].first - dis[i].first) + square(dis[x].second - dis[i].second) <= square(d))able[i][x] = 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;
        }
    }
    return 0;
}