并查集+Set-BZOJ-1604-[Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
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Description
了解奶牛们的人都知道,奶牛喜欢成群结队.观察约翰的N(1≤N≤100000)只奶牛,你会发现她们已经结成了几个“群”.每只奶牛在吃草的时候有一个独一无二的位置坐标Xi,Yi(l≤Xi,Yi≤[1..10^9];Xi,Yi∈整数.当满足下列两个条件之一,两只奶牛i和j是属于同一个群的:
1.两只奶牛的曼哈顿距离不超过C(1≤C≤10^9),即lXi - xil+lYi - yil≤C.
2.两只奶牛有共同的邻居.即,存在一只奶牛k,使i与k,j与k均同属一个群.
给出奶牛们的位置,请计算草原上有多少个牛群,以及最大的牛群里有多少奶牛
Input
第1行输入N和C,之后N行每行输入一只奶牛的坐标.
Output
仅一行,先输出牛群数,再输出最大牛群里的牛数,用空格隔开.
Sample Input
4 2
1 1
3 3
2 2
10 10
Line 1: A single line with a two space-separated integers: the number of cow neighborhoods and the size of the largest cow neighborhood.
Sample Output
2 3
OUTPUT DETAILS:
There are 2 neighborhoods, one formed by the first three cows andthe other being the last cow. The largest neighborhood thereforehas size 3.
题解:
很早就开了这道题,一直没有A掉,今天课上无聊补了一发相关姿势,还是学了老半天才会,1A。
一般遇到这种二维的距离题,都可以在数学上做文章,此题用到曼哈顿距离,那么我们把公式|Xi-xi|+|Yi-yi|拆开,成了下面4个表达式。
(1)Xi-xi+Yi-yi
(2)Xi-xi+yi-Yi
(3)xi-Xi+Yi-yi
(4)xi-Xi+yi-Yi
那么其实可以将(1)(4)式合并为|(Xi+Yi)-(xi+yi)|,把(2)(3)式合并为|(Xi-Yi)-(xi-yi)|。
我们用A来代替x+y,B来代替x-y,则其实只要max(Ai-ai,Bi-bi)<=c就满足题目中距离要求了。
那么我们对于每个点(牛),都用A,B去表示。只需要先整体以A大小为标准进行一次排序,然后维护一个队列,使得队列中队首的A与队尾的A之差小于c即可,这样保证了队列中的点都是满足delta(A)不大于c的。同时利用set的平衡二叉树高效查询,维护队列中点的B值,遍历到一个点时,先得到满足当前情况的队列,然后得出以该点的B应该处于set中哪个位置,并与前后的B进行比较,满足delta(B)不大于c的就合并。
最后扫一下father数组就可以得答案了。
#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <set>#include <vector>#define MAXN 100005#define INF 0x3f3f3f3fusing namespace std;long long int c,n;long long int father[MAXN];long long int total,head;typedef struct node{ long long int x,y; bool operator <(const node&r)const { return x<r.x; }}node;multiset<node>S;multiset<node>::iterator it;node N[MAXN];long long int find(long long int num){ long long int root,tmp,now=num; while(father[now]!=now) { now=father[now]; } root=now; now=num; while(father[now]!=now) { tmp=father[now]; father[now]=root; now=tmp; } return root;}void unit(long long int a,long long int b){ long long int fa=find(a),fb=find(b); if(fa!=fb) father[fa]=fb;}long long int num[MAXN];int main(){ long long int x,y; cin >> n >> c; for(long long int i=1;i<=n;i++) { scanf("%lld %lld",&x,&y); N[i].x=x+y,N[i].y=x-y; father[i]=i; } sort(N+1,N+n+1); S.insert(node{-INF,0}),S.insert(node{INF,0}); head=1; for(long long int i=1;i<=n;i++) { while(N[i].x-N[head].x>c) { it=S.find(node{N[head].y,head}); S.erase(it); head++; } it=S.lower_bound(node{N[i].y,i}); node tmp1,tmp2; tmp1=*it; tmp2=*(--it); if(tmp1.x-N[i].y<=c) unit(tmp1.y,i); if(N[i].y-tmp2.x<=c) unit(tmp2.y,i); S.insert(node{N[i].y,i}); } long long int out1=0,out2=0; for(long long int i=1;i<=n;i++) num[find(i)]++; for(long long int i=1;i<=n;i++) { if(num[i]) { out1++; out2=max(out2,num[i]); } } cout << out1 << " " << out2 << endl; return 0;}
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