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
", ans);
	}
	return 0;
}