POJ

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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<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<vector>#include<iostream>using namespace std;#define INF 0x3f3f3f3fconst 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<Edge> 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;}