zoj 3717 Balloon 2-sat

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。

分析：

　　二分的思想很明显，二分答案之后，建图：如果两点之间的距离小于二分值时，连接相应的边，通过2-sat判断一下即可。

　　注意到题目的要求是向下取整，于是我们可以先*10000，最后直接取模即可。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define debug puts("here")
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)
#define pb push_back
#define RD(n) scanf("%d",&n)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)
#define All(vec) vec.begin(),vec.end()
#define MP make_pair
#define PII pair<int,int>
#define PQ priority_queue
#define cmax(x,y) x = max(x,y)
#define cmin(x,y) x = min(x,y)
#define Clear(x) memset(x,0,sizeof(x))

/******** program ********************/
const int MAXN = 405;
const int MAXM = 1000005;

int low[MAXN],sta[MAXN],dfn[MAXN],fa[MAXN],dep,bcnt,top;
bool use[MAXN];
int po[MAXN],tol,n,m;

struct Edge {
    int y,next;
} edge[MAXM];

struct Point {
    int a[3];
    void rd() {
        RD3(a[0],a[1],a[2]);
        a[0] *= 10000;
        a[1] *= 10000;
        a[2] *= 10000;
    }
} p[MAXN][2];

void add(int x,int y) {
    edge[++tol].y = y;
    edge[tol].next = po[x];
    po[x] = tol;
}

void dfs(int x) {
    sta[++top] = x;
    use[x] = true;
    low[x] = dfn[x] = ++ dep;
    int y;
    for(int i=po[x]; i; i=edge[i].next) {
        y = edge[i].y;
        if(!dfn[y]) {
            dfs(y);
            low[x] = min(low[x],low[y]);
        } else if(use[y])
            low[x] = min(low[x],dfn[y]);
    }
    if(low[x]==dfn[x]) {
        ++ bcnt;
        do {
            y = sta[top--];
            use[y] = false;
            fa[y] = bcnt;
        } while(x!=y);
    }
}

bool sat() {
    memset(dfn,0,sizeof(dfn));
    memset(use,false,sizeof(use));
    dep = top = bcnt = 0;
    rep1(i,2*n)
    if(!dfn[i])
        dfs(i);
    rep1(i,n)
    if(fa[i]==fa[i+n])
        return false;
    return true;
}

ll sqr(ll x) {
    return x*x;
}

ll dist(Point a,Point b) {
    return sqr(a.a[0]-b.a[0])+sqr(a.a[1]-b.a[1])+sqr(a.a[2]-b.a[2]);
}

bool solve(ll mid) {
    memset(po,0,sizeof(po));
    tol = 0;
    rep1(i,n) {
        rep1(j,n) {
            if(i==j)continue;
            ll d = dist(p[i][0],p[j][0]);
            if(d<mid) {
                add(i,j+n);
                add(j,i+n);
            }

            d = dist(p[i][0],p[j][1]);
            if(d<mid) {
                add(i,j);
                add(j+n,i+n);
            }

            d = dist(p[i][1],p[j][0]);
            if(d<mid) {
                add(i+n,j+n);
                add(j,i);
            }

            d = dist(p[i][1],p[j][1]);
            if(d<mid) {
                add(i+n,j);
                add(j+n,i);
            }
        }
    }
    return sat();
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("sum.in", "r", stdin);
    // freopen("cf.out", "w", stdout);
#endif

    while(~RD(n)) {
        rep1(i,n) {
            p[i][0].rd();
            p[i][1].rd();
        }
        ll ans = -1 , l = 0 , r = 2000000000LL;
        while(l<=r) {
            ll mid = (l+r)/2;
            if(solve(mid*mid)) {
                l = mid+1;
                ans = mid;
            } else r = mid-1;
        }
        printf("%lld.%03lld
",ans/20000,(ans/2%10000)/10 );
    }

    return 0;
}