ZOJ3717 Balloon(2-SAT)

一个很玄乎的问题，但听到2-SAT之后就豁然开朗了。题目的意思是这样的，给你n个点群，每个点群里面有两个点，你要在每个点群里面选一个点，以这些点做半径为r的圆，然后r会有一个最大值，问的就是怎么选这些点使得r最大。

2-SAT就是对于每个变量有一些制约的关系   a->b 表示选了a就就要选b。然后我们二分这个半径，对于两点间距离<2*r的点（a,b）选了a就不能选b,选了b就不能选a，以此构图。然后跑一次强连通分量。最后判是否有解的时候就是判对于两个属于相同点群的点，它们不能处于同一强连通分量下。写的时候跪的点实在太多了，数组越界呀，强连通写错呀，精度呀，这样的题太坑爹了－ －0

#pragma warning(disable:4996)

#include<iostream>

#include<cstring>

#include<cstdio>

#include<algorithm>

#include<cmath>

#include<vector>

#define ll long long

#define maxn 220

#define eps 1e-8

using namespace std;



struct Point

{

	double x, y, z;

	Point(double xi, double yi, double zi) :x(xi), y(yi), z(zi){}

	Point(){}

}p[maxn * 2];



double dist(Point a, Point b){

	return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y) + (a.z - b.z)*(a.z - b.z));

}



double dis[maxn * 2][maxn * 2];



int low[maxn * 2];

int pre[maxn * 2];

int dfs_clock;

int sta[maxn * 2];

int st;

int sccno[maxn * 2];

int n;

vector<int> G[maxn * 2];

int scc_cnt;



int dcmp(double x){

	return (x > eps) - (x < -eps);

}



void dfs(int u){

	low[u] = pre[u] = ++dfs_clock;

	sta[++st] = u;

	for (int i = 0; i < G[u].size(); i++){

		int v = G[u][i];

		if (!pre[v]){

			dfs(v);

			low[u] = min(low[u], low[v]);

		}

		else if (!sccno[v]){

			low[u] = min(low[u], pre[v]);

		}

	}

	if (low[u] == pre[u]){

		++scc_cnt;

		while (1){

			int x = sta[st]; st--;

			sccno[x] = scc_cnt;

			if (x == u) break;

		}

	}

}



bool judge(double x)

{

	memset(sccno, 0, sizeof(sccno));

	memset(pre, 0, sizeof(pre));

	memset(low, 0, sizeof(low));

	st = 0; dfs_clock = 0;

	scc_cnt = 0;

	for (int i = 0; i <= 2 * n; i++) G[i].clear();



	for (int i = 0; i < n; i++){

		for (int j = i + 1; j < n; j++){

			if (dcmp(dis[i][j] - 2 * x) < 0){

				G[i].push_back(j + n);

				G[j].push_back(i + n);

			}

			if (dcmp(dis[i][j + n] - 2 * x) < 0){

				G[i].push_back(j);

				G[j + n].push_back(i + n);

			}

			if (dcmp(dis[i + n][j] - 2 * x) < 0){

				G[i + n].push_back(j + n);

				G[j].push_back(i);

			}

			if (dcmp(dis[i + n][j + n] - 2 * x) < 0){

				G[i + n].push_back(j);

				G[j + n].push_back(i);

			}

		}

	}

	for (int i = 0; i < 2 * n; i++){

		if (!pre[i]) dfs(i);

	}

	for (int i = 0; i < n; i++){

		if (sccno[i] == sccno[i + n]) return false;

	}

	return true;

}



int main()

{

	while (cin >> n)

	{

		for (int i = 0; i < n; i++){

			scanf("%lf%lf%lf", &p[i].x, &p[i].y, &p[i].z);

			scanf("%lf%lf%lf", &p[i + n].x, &p[i + n].y, &p[i + n].z);

		}

		for (int i = 0; i < 2 * n; i++){

			for (int j = i + 1; j < 2 * n; j++){

				dis[i][j] = dis[j][i] = dist(p[i], p[j]);

			}

		}

		double l = 0, r = 1e10;

		while (dcmp(r - l)>0){

			double mid = (l + r) / 2;

			if (judge(mid)) l = mid;

			else r = mid;

		}

		int tmp = l * 1000;

		double ans = tmp / 1000.0;

		printf("%.3lf\n", ans);

	}

	return 0;

}