|Hdu 3622|2-SAT|二分|Bomb Game

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/*    hdu 3622     二分后2-SAT判断    本题教训：    1、2-SAT加边视情况加    2、浮点数二分查找的写法     3、浮点数的eps最好开多一点，防卡精度 */#include<cstdio>#include<cstring>#include<vector>#include<cmath>#include<algorithm>#define ll long long#define db double#define ms(i,j) memset(i,j)#define INF 100000000#define llINF 1LL<<60using namespace std;const int MAXN = (100 + 5) * 2;struct TwoSat{    int mark[MAXN*2];    int S[MAXN*2];    vector<int> G[MAXN*2];    int n, c;    int init(int ni)    {        n = ni;        for (int i=0;i<=n*2;i++)        {            mark[i] = false;            G[i].clear();        }    }    int dfs(int u)    {        if (mark[u^1]) return false;        if (mark[u]) return true;        mark[u] = true;        S[++c] = u;        for (int i=0;i<G[u].size();i++)        {            if (!dfs(G[u][i])) return false;        }        return true;    }    int 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;    }}ts;int n;int xi[MAXN], yi[MAXN];db dis(int x, int y){    return sqrt(1.0*(xi[x]-xi[y])*(xi[x]-xi[y]) + (yi[x]-yi[y])*(yi[x]-yi[y]));}int init(){    for (int i=0;i<n;i++)    {        scanf("%d%d%d%d", &xi[i], &yi[i], &xi[i+n], &yi[i+n]);    }}int solve(){    db l = 0.0, r = 40000.0;    db ans = 0;    for (int i=0;i<100;i++)    {        db mid = (l+r)/2.0;        ts.init(n);        for (int i=0;i<n;i++)        for (int j=i+1;j<n;j++)        {            int x = i*2, y = j*2;            if (dis(i,j)+0.0000001<mid*2)            {                ts.G[x].push_back(y^1);                ts.G[y].push_back(x^1);            }            if (dis(i,j+n)+0.0000001<mid*2)            {                ts.G[x].push_back(y);                ts.G[y^1].push_back(x^1);            }            if (dis(i+n,j)+0.0000001<mid*2)            {                ts.G[x^1].push_back(y^1);                ts.G[y].push_back(x);            }            if (dis(i+n,j+n)+0.0000001<mid*2)            {                ts.G[x^1].push_back(y);                ts.G[y^1].push_back(x);            }        }        if (ts.solve())         {            ans = l = mid;        } else r = mid;    }    printf("%.2f\n", ans);}int main(){    while (scanf("%d", &n)==1)    {        init();        solve();    }    return 0;}
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