BZOJ200题纪念!

一个非常巧妙的方法求曼哈顿距离:

如果原来坐标是(x, y),令新的坐标为(X, Y), 其中X = x + y, Y = x - y

那么:曼哈顿距离 = |x1 - x2| + |y1 - y2| = max(|X1 - X2|, |Y1 - Y2|)

于是我们先进行坐标变换,按X排序。

然后用一个队列来做,满足队尾X - 队首X < c。

对这个队列中每个点的Y维护一棵平衡树,如果新加入元素的前驱后继与它的Y值差值不超过c,则用并查集将他们连在一起。

(其实就是类似kruskal的改进版本)

蒟蒻不会平衡树,于是又使用了STL的multiset。。。

 

 1 /**************************************************************
 2     Problem: 1604
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:868 ms
 7     Memory:5396 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <algorithm>
12 #include <set>
13  
14 using namespace std;
15 typedef long long ll;
16 const ll inf = (ll) 1e16;
17 const int N = 100005;
18  
19 struct data{
20     ll x, y;
21     int w;
22 }a[N];
23 inline bool operator < (const data &a, const data &b){
24     return a.y < b.y;
25 }
26 inline bool cmpx (const data &a, const data &b){
27     return a.x < b.x;
28 }
29  
30 int fa[N], cnt[N], ans, ANS;
31 int n, c, X, Y;
32 multiset <data> S;
33 set <data> ::iterator it;
34  
35 int x;
36 char ch;
37 inline unsigned int read(){
38     x = 0;
39     ch = getchar();
40     while (ch < '0' || ch > '9')
41         ch = getchar();
42          
43     while (ch >= '0' && ch <= '9'){
44         x = x * 10 + ch - '0';
45         ch = getchar();
46     }
47     return x;
48 }
49  
50 int find_fa(int x){
51     return fa[x] == x ? x : fa[x] = find_fa(fa[x]);
52 }
53  
54 inline void Union(int x, int y){
55     x = find_fa(x), y = find_fa(y);
56     if (x != y)
57         fa[x] = y, --ans;
58 }
59  
60 void work(){
61     for (int i = 1; i <= n; ++i)
62         fa[i] = i;
63     S.insert((data) {0, inf, 0});
64     S.insert((data) {0, -inf, 0});
65     S.insert(a[1]);
66     int now = 1;
67     data l, r;
68     for (int i = 2; i <= n; ++i){
69         while (a[i].x - a[now].x > c)
70             S.erase(S.find(a[now++]));
71         it = S.lower_bound(a[i]);
72         r = *it, l = *--it;
73         if (a[i].y - l.y <= c)
74             Union(a[i].w, l.w);
75         if (r.y - a[i].y <= c)
76             Union(a[i].w, r.w);
77         S.insert(a[i]);
78     }
79 }
80  
81 int main(){
82     n = read(), c = read(), ans = n;
83     for (int i = 1; i <= n; ++i){
84         X = read(), Y = read();
85         a[i].x = X + Y, a[i].y = X - Y, a[i].w = i;
86     }
87     sort(a + 1, a + n + 1, cmpx);
88      
89     work();
90     for (int i = 1; i <= n; ++i)
91         ++cnt[find_fa(i)];
92     for (int i = 1; i <= n; ++i)
93         ANS = max(ANS, cnt[i]);
94     printf("%d %d\n", ans, ANS);
95     return 0;
96 }
View Code

 

posted on 2014-10-22 11:03  Xs酱~  阅读(607)  评论(0编辑  收藏  举报