ZOJ 3717 Balloon（2-sat）

Balloon

Time Limit: 3 Seconds Memory Limit: 65536 KB Special Judge

The weather is wonderful today. Gao takes a walk in the garden with his girlfriend. His girlfriend likes balloons so much, so that she wants to fly some balloons (not kites!)
in the garden.
We can regard the garden as a three-dimensional space, and its coordinate starts from (0,0,0) to (10000,10000,10000). There are N groups of balloons, each of groups has a red
balloon and a blue balloon. We can regard each balloon as a sphere, all the radius of spheres are R. The center of each sphere will be given in the input.
For some reasons, she wants to choose one balloon (red one or blue one) from each group, so that she can put exactly N balloons in the garden. It's obvious
that there is no overlap for any two balloons in the N balloons which she selected. The largest R will make Gao's girlfriend happiest. Can you help Gao to calculate the largest R?

Input

There are multiple cases. For each case, The first line contains an integer N (2 ≤ N ≤ 200), meaning there are N groups of balloons. In the next N lines,
each line contains six integers, indicating the coordinates of two balloons.
NOTICE: The garden only limits the center of the balloon.

Output

For each test case, it contains one real number indicating the largest R. The results should be rounded to three decimal places. You should promise that there is still no overlap
for any two balloons after rounded.

Sample Input

2
1 1 1 5 5 5
1 1 1 5 5 5

Sample Output

3.464

Author: REN, Qing

题目：给出空间中的n对点，要求从每对点中选出一个，使得最近的点的距离最远。

题解：2-sat 模板，二分答案。

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#define Mod 1000000007
#define ll long long

using namespace std;

const int MAXN = 1010;
const int MAXM = 500010;

int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
int Index,top;
int scc,n;
int head[MAXN],tot;
bool Instack[MAXN];
int num[MAXN];
struct Edge {
    int to,next;
} edge[MAXM];

struct Point {
    double x,y,z;
} p[MAXN][2];

void init() {
    tot = 0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v) {
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}

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));
}

void Tarjan(int u) {
    int v;
    Low[u] = DFN[u] = ++Index;
    Stack[top++] = u;
    Instack[u] = true;
    for(int i = head[u]; i != -1; i = edge[i].next) {
        v = edge[i].to;
        if( !DFN[v] ) {
            Tarjan(v);
            if(Low[u] > Low[v])Low[u] = Low[v];
        } else if(Instack[v] && Low[u] > DFN[v])
            Low[u] = DFN[v];
    }
    if(Low[u] == DFN[u]) {
        scc++;
        do {
            v = Stack[--top];
            Instack[v] = false;
            Belong[v] = scc;
            num[scc]++;
        } while(v != u);
    }
}

bool solve(int n) {
    memset(DFN,0,sizeof(DFN));
    memset(Instack,false,sizeof(Instack));
    memset(num,0,sizeof(num));
    Index = scc = top = 0;
    for(int i = 0; i < n; i++)
        if(!DFN[i])
            Tarjan(i);
    for(int i = 0; i < n; i += 2) {
        if(Belong[i] == Belong[i^1])
            return false;
    }
    return true;
}

bool C(double R) {
    init();
    for(int i=0; i<n; i++) {
        for(int j=i+1; j<n; j++) {
            for(int k=0; k<2; k++) {
                for(int l=0; l<2; l++) {
                    if(dist(p[i][k],p[j][l])<=2*R) {
                        addedge(i*2+k,j*2+(l^1));
                        addedge(j*2+l,i*2+(k^1));
                    }
                }
            }
        }
    }
    return solve(2*n);
}

int main() {
    //freopen("test.in","r",stdin);
    while(cin>>n) {
        for(int i=0; i<n; i++) {
            for(int j=0; j<2; j++) {
                scanf("%lf%lf%lf",&p[i][j].x,&p[i][j].y,&p[i][j].z);
            }
        }
        double l=0.0,r=1000000000000.0+1;
        for(int i=0; i<100; i++) {
            double mid=(l+r)/2;
            if(C(mid))l=mid;
            else r=mid;
        }
        printf("%.3f\n",l-0.0005);
    }
    return 0;
}