ZOJ 3717 Balloon ( TLE )

正解2-SAT。

我用DLX想搜一搜的，结果TLE了……

没什么遗憾，最起码我尝试过了。

扔个代码留作纪念。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>

using namespace std;

const int INF =  << ;
const int MAXN = ;
const double eps = 1e-;

struct Point
{
    double x, y, z;
    Point( double x = , double y = , double z =  ):x(x), y(y), z(z){ }
    void readPoint()
    {
        scanf( "%lf%lf%lf", &x, &y, &z );
        return;
    }
};

bool mx[MAXN][];    //01矩阵
int C[MAXN*], cnt[];
int U[MAXN*], D[MAXN*];
int L[MAXN*], R[MAXN*];
int head;
int maxr, maxc;
Point P[MAXN];
int N;

int dcmp( double a )
{
    if ( fabs(a) < eps ) return ;
    return a <  ? - : ;
}

double Dis( 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 Remove( int c )
{
    int i, j;
    L[ R[c] ] = L[c];
    R[ L[c] ] = R[c];
    for ( i = D[c]; i != c; i = D[i] )
    {
        for ( j = R[i]; j != i; j = R[j] )
        {
            U[ D[j] ] = U[j];
            D[ U[j] ] = D[j];
            --cnt[ C[j] ];
        }
    }
    return;
}

void Resume( int c )
{
    int i, j;
    R[ L[c] ] = c;
    L[ R[c] ] = c;
    for ( i = D[c]; i != c; i = D[i] )
    {
        for ( j = R[i]; j != i; j = R[j] )
        {
            U[ D[j] ] = j;
            D[ U[j] ] = j;
            ++cnt[ C[j] ];
        }
    }
    return;
}

bool DFS( int cur )
{
    int i, j, c, minv;

    if ( cur > N ) return false;
    if ( R[head] == head )
    {
        if ( cur == N )
            return true;
        return false;
    }

    minv = INF;
    for ( i = R[head]; i != head; i = R[i] )
    {
        if ( cnt[i] < minv )
        {
            minv = cnt[i];
            c = i;
        }
    }

    Remove(c);
    for ( i = D[c]; i != c; i = D[i] )
    {
        for( j = R[i]; j != i; j = R[j] )
            Remove( C[j] );

        if ( DFS( cur +  ) ) return true;

        for( j = R[i]; j != i; j = R[j] )
            Resume( C[j] );
    }

    Resume(c);
    return false;
}

bool build()
{
    int i, j, cur, pre, first;
    head = ;
    for ( i = ; i < maxc; ++i )
    {
        R[i] = i + ;
        L[i + ] = i;
    }
    R[ maxc ] = ;
    L[] = maxc;

    //列双向链表
    for ( j = ; j <= maxc; ++j )
    {
        pre = j;
        cnt[j] = ;
        for ( i = ; i <= maxr; ++i )
        {
            if ( mx[i][j] )
            {
                ++cnt[j];
                cur = i * maxc + j;  //当前节点的编号
                C[cur] = j;          //当前节点所在列
                D[pre] = cur;
                U[cur] = pre;
                pre = cur;
            }
        }
        U[j] = pre;
        D[pre] = j;
        if ( !cnt[j] ) return false; //一定无解
    }

    //行双向链表
    for ( i = ; i <= maxr; ++i )
    {
        pre = first = -;
        for ( j = ; j <= maxc; ++j )
        {
            if( mx[i][j] )
            {
                cur = i * maxc + j;
                if ( pre == - ) first = cur;
                else
                {
                    R[pre] = cur;
                    L[cur] = pre;
                }
                pre = cur;
            }
        }
        if ( first != - )
        {
            R[pre] = first;
            L[first] = pre;
        }
    }
    return true;
}

/**************以上DLX模板*****************/

void show()
{
    for ( int i = ; i <= maxr; ++i )
    {
        for ( int j = ; j <= maxc; ++j )
            printf( "%d", mx[i][j] );
        puts("");
    }
    puts("---------");
    return;
}

//得到该情况下的01矩阵
void init( double R )
{
    memset( mx, false, sizeof(mx) );

    for ( int i = ; i <= maxr; ++i )
    {
        mx[i][i] = true;
        if ( i %  )
        {
            //printf("ii %d %d\n", i, i + 1);
            mx[i][N + N + i/+] = true;
            mx[i + ][N + N + i/+] = true;
        }
        for ( int j = ; j <= maxr; ++j )
        {
            //printf("i=%d j=%d\n", i, j );
            if ( dcmp( Dis( P[i], P[j] ) - 2.0 * R ) <  )
                mx[j][i] = true;
        }
    }
    //show();
    return;
}

bool ok( double mid )
{
    init( mid );
    if ( build() == false ) return false;
    if ( DFS(  ) == false ) return false;
    return true;
}

double solved()
{
    double l = ;
    double r = Dis( Point(, , ), Point( , ,  ) );
    double ans;
    while ( dcmp( r - l ) >  )
    {
        double mid = ( l + r ) / 2.0;
        //printf( "mid = %.6f\n", mid );
        if ( ok( mid ) )
        {
            l = mid;
            ans = mid;
        }
        else r = mid;
    }
    return ans;
}

int main()
{
    //freopen( "in.txt", "r", stdin );
    //freopen( "out.txt", "w", stdout );
    while ( scanf( "%d", &N ) ==  )
    {
        maxr = ;
        for ( int i = ; i < N; ++i )
        {
            P[++maxr].readPoint();
            P[++maxr].readPoint();
        }
        maxc = maxr + N;
        double ans=(int)(solved()*+0.5+eps)/1000.0;
        if ( !ok(ans) ) ans -= 0.001;
        printf( "%.3f\n", ans );
    }
    return ;
}