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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 }
BZOJ1604 [Usaco2008 Open]Cow Neighborhoods 奶牛的邻居
标签:style blog http color io os ar 使用 for
原文地址:http://www.cnblogs.com/rausen/p/4042565.html