poj3241 曼哈顿最小生成树

题目：求平面上n个点的曼哈顿最小生成树的第k大的边。

思路：模板题

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3f// 0x3f3f3f3fconst int maxn=1e4+50;const int maxm=2e3+50;int tree[maxm],pos[maxm];int lowbit(int i){    return i&(-i);}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;}struct Node{    int id,x,y,xsy;}point[maxn],p[maxn];bool cmpx(Node a,Node b){    if(a.x==b.x) return a.y>b.y;    return a.x>b.x;}bool cmpxsy(Node a,Node b){    return a.xsy<b.xsy;}struct Edge{    int u,v,w;    Edge(int _u=0,int _v=0,int _w=0):u(_u),v(_v),w(_w){}    bool operator<(const Edge& rhs)const{        return w<rhs.w;    }};vector<Edge> edge;int dist(Node a,Node b){    return f_abs(a.x-b.x)+f_abs(a.y-b.y);}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) edge.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){    edge.clear();    for(int i=0;i<4;i++){        for(int j=0;j<n;j++) point[j]=p[j];        for(int j=0;j<n;j++){            if(i==1) swap(point[j].x,point[j].y);            else if(i==2) point[j].y=-point[j].y;            else if(i==3){                swap(point[j].x,point[j].y);                point[j].y=-point[j].y;            }        }        Manhaton(n);    }}int fa[maxn];int findset(int x){    if(x==fa[x]) return x;    return fa[x]=findset(fa[x]);}int MST(int n,int k){    if(n<=k) return 0;    for(int i=0;i<n;i++) fa[i]=i;    sort(edge.begin(),edge.end());    int sz=edge.size();    for(int i=0;i<sz;i++){        int u=findset(edge[i].u);        int v=findset(edge[i].v);        if(u==v) continue;        fa[u]=v;        n--;        if(n==k) return edge[i].w;    }    return 0;}int main(){    int n,k;    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;}