POJ - 3241 Object Clustering(莫队算法/曼哈顿最小生成树)

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题意:给出n个点,第i个点的坐标为(xi,yi),求这n个点形成的曼哈顿最小生成树的第k大边.

分析:曼哈顿最小生成树的模板题(平面曼哈顿最小生成树的详解

参考代码:

#include
#include
#include
#include
#include
#include

using namespace std;
#define INF 0x3f3f3f3f
const int maxn = 1e4+10;
const int maxm = 2e3+10;
int n,k;
int tree[maxm],pos[maxm];
struct Point{
    int x,y,id;
    int xsy;
};
Point point[maxn],p[maxn];
struct Edge{
    int u,v,w;
    Edge(){}
    Edge( int uu, int vv, int ww)
    {
        u = uu, v = vv, w = ww;
    }
    bool operator < ( const Edge &e)const
    {
        return w < e.w;
    }
};
vector e;
int fa[maxn];

bool cmpx( Point p1, Point p2)
{
    if( p1.x != p2.x)
        return p1.x > p2.x;
    return p1.y > p2.y;
}

bool cmpxsy( Point p1, Point p2)
{
    return p1.xsy < p2.xsy;
}

int dist( Point p1, Point p2)
{
    return abs(p1.x-p2.x)+abs(p1.y-p2.y);
}

inline int lowbit( int x)
{
    return x&-x;
}

void add( int i, int val, int id)
{
    while( i < maxm)
    {
        if( tree[i] > val)
        {
            tree[i] = val;
            pos[i] = id;
        }
        i += lowbit(i);
    }
}

int query( int i)
{
    int id = -1, val = INF;
    while( i > 0)
    {
        if( tree[i] < val)
        {
            val = tree[i];
            id = pos[i];
        }
        i -= lowbit(i);
    }
    return id;
}

void Manhaton( int n)
{
    for( int i = 0; i < n; i++)
        point[i].xsy = point[i].x-point[i].y;
    sort(point,point+n,cmpxsy);
    int now = 0, pre = -INF;
    for( int i = 0; i < n; i++)
    {
        if( pre != point[i].xsy)
        {
            now++;
            pre = point[i].xsy;
        }
        point[i].xsy = now;
    }
    sort(point,point+n,cmpx);
    for( int i = 1; i < maxm; i++)
        tree[i] = INF, pos[i] = -1;
    for( int i = 0; i < n; i++)
    {
        int u = point[i].id;
        int v = query(point[i].xsy);
        if( v != -1)
            e.push_back(Edge(u,v,dist(p[u],p[v])));
        add(point[i].xsy,point[i].x+point[i].y,u);
    }
}

void BuildEdge( int n)
{
    e.clear();
    for( int cnt = 0; cnt < 4; cnt++)
    {
        for( int i = 0; i < n; i++)
            point[i] = p[i];
        for( int i = 0; i < n; i++)
        {
            if(cnt == 1)
                swap(point[i].x,point[i].y);
            else if( cnt == 2)
                point[i].y = -point[i].y;
            else if( cnt == 3)
            {
                swap(point[i].x,point[i].y);
                point[i].y = -point[i].y;
            }
        }
        Manhaton(n);
    }
}

int find( int x)
{
    if( x == fa[x])
        return x;
    return fa[x] = find(fa[x]);
}

int MST( int n, int k)
{
    if( n <= k)
        return 0;
    for( int i = 0; i < n; i++)
        fa[i] = i;
    sort(e.begin(),e.end());
    int sz = e.size();
    for( int i = 0; i < sz; i++)
    {
        int u = find(e[i].u);
        int v = find(e[i].v);
        if( u == v)
            continue;
        fa[u] = v;
        n--;
        if( n == k)
            return e[i].w;
    }
}

int main()
{
    while( ~scanf("%d%d",&n,&k))
    {
        for( int i = 0; i < n; i++)
        {
            scanf("%d%d",&p[i].x,&p[i].y);
            p[i].id = i;
        }
        BuildEdge(n);
        printf("%d\n",MST(n,k));
    }

    return 0;
}