zoj 3717 - Balloon(2-SAT)

裸的2-SAT，详见刘汝佳训练指南P-323

不过此题有个特别需要注意的地方：You should promise that there is still no overlap for any two balloons after rounded.

模版题，

代码如下：

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <string>#include <cmath>#include <queue>#include <stack>#include <vector>#include <map>#define LL long long#define eps 1e-5#define M 205#define mod 1000000007using namespace std;struct Point{    int x, y, z;    Point() { }    Point(int x, int y, int z) : x(x), y(y), z(z) { }    void readPoint()    {        scanf("%d %d %d", &x, &y, &z);    }};struct TwoSAT{    int n;    vector<int>G[M*2];    bool mark[M*2];    int S[M*2], c;    bool dfs(int x)    {        if(mark[x^1]) return false;        if(mark[x]) return true;        mark[x] = true;        S[c++] = x;        for(int i = 0; i < (int)G[x].size(); ++i)            if(!dfs(G[x][i])) return false;        return true;    }    void init(int n)    {        this->n = n;        for(int i = 0; i < n*2; ++i)    G[i].clear();        memset(mark, 0, sizeof(mark));    }    void add_clause(int x, int y)    {        G[x^1].push_back(y);        G[y^1].push_back(x);    }    bool solve()    {        for(int i = 0; i < n*2; i+=2)            if(!mark[i] && !mark[i+1])            {                c = 0;                if(!dfs(i))                {                    while(c>0) mark[S[--c]] = false;                    if(!dfs(i+1)) return false;                }            }        return true;    }};Point poi[2*M];int dcmp(double a){    if(fabs(a)<eps) return 0;    return a<0?-1:1;}double dis(Point a, Point b){    return sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y)+1.0*(a.z-b.z)*(a.z-b.z));}int ok(int n, double R){    TwoSAT temp;    temp.init(n);    for(int i = 0; i < 2*n; ++i)        for(int j = i+1; j < 2*n; ++j)            if(dcmp(dis(poi[i], poi[j])-2*R)<0)            {                temp.add_clause(i^1, j^1);            }    return temp.solve();}int main (){    int n;    while(~scanf("%d", &n))    {        for(int i = 0; i < 2*n; ++i)            poi[i].readPoint();        double l = 0.0, r = dis(Point(0,0,0), Point(10000,10000,10000)), mid;        while(r-l>eps)        {            mid = (r+l)/2;            if(ok(n, mid))                l = mid;            else                r = mid;        }        double ans = (int)(mid*1000+0.5)/1000.0;//注意！！！        if(!ok(n, ans)) ans-=0.001;        printf("%.3lf\n", ans);    }    return 0;}