POJ 3241 Object Clustering（Manhattan MST）

题目链接：http://poj.org/problem?id=3241

Description

We have N (N ≤ 10000) objects, and wish to classify them into several groups by judgement of their resemblance. To simply the model, each object has 2 indexes a and b (a, b ≤ 500). The resemblance of object i and object j is defined by dij = |ai - aj| + |bi - bj|, and then we say i is dij resemble to j. Now we want to find the minimum value of X, so that we can classify the N objects into K (K < N) groups, and in each group, one object is at most X resemble to another object in the same group, i.e, for every object i, if i is not the only member of the group, then there exists one object j (i ≠ j) in the same group that satisfies dij ≤ X

Input

The first line contains two integers N and K. The following N lines each contain two integers a and b, which describe a object.

Output

A single line contains the minimum X.

 

题目大意：给n个点，两个点之间的距离为曼哈顿距离。要求把n个点分成k份，每份构成一个连通的子图（树），要求最大边最小。求最大边的最小值。

思路：实际上就是求平面上曼哈顿距离的最小生成树的第k大边（即减掉最大的k-1条边构成k份）。

资料：曼哈顿MST。复杂度O(nlogn)。

http://wenku.baidu.com/view/1e4878196bd97f192279e941.html

http://blog.csdn.net/huzecong/article/details/8576908

 

代码（79MS）：

  1 #include <cstdio>
  2 #include <iostream>
  3 #include <cstring>
  4 #include <algorithm>
  5 using namespace std;
  6 typedef long long LL;
  7 #define FOR(i, n) for(int i = 0; i < n; ++i)
  8 
  9 namespace Bilibili {
 10     const int MAXV = 10010;
 11     const int MAXE = MAXV * 4;
 12 
 13     struct Edge {
 14         int u, v, cost;
 15         Edge(int u = 0, int v = 0, int cost = 0):
 16             u(u), v(v), cost(cost) {}
 17         bool operator < (const Edge &rhs) const {
 18             return cost < rhs.cost;
 19         }
 20     };
 21 
 22     struct Point {
 23         int x, y, id;
 24         void read(int i) {
 25             id = i;
 26             scanf("%d%d", &x, &y);
 27         }
 28         bool operator < (const Point &rhs) const {
 29             if(x != rhs.x) return x < rhs.x;
 30             return y < rhs.y;
 31         }
 32     };
 33 
 34     Point p[MAXV];
 35     Edge edge[MAXE];
 36     int x_plus_y[MAXV], y_sub_x[MAXV];
 37     int n, k, ecnt;
 38 
 39     int hash[MAXV], hcnt;
 40 
 41     void get_y_sub_x() {
 42         for(int i = 0; i < n; ++i) hash[i] = y_sub_x[i] = p[i].y - p[i].x;
 43         sort(hash, hash + n);
 44         hcnt = unique(hash, hash + n) - hash;
 45         for(int i = 0; i < n; ++i) y_sub_x[i] = lower_bound(hash, hash + hcnt, y_sub_x[i]) - hash + 1;
 46     }
 47 
 48     void get_x_plus_y() {
 49         for(int i = 0; i < n; ++i) x_plus_y[i] = p[i].x + p[i].y;
 50     }
 51 
 52     int tree[MAXV];
 53     int lowbit(int x) {
 54         return x & -x;
 55     }
 56 
 57     void update_min(int &a, int b) {
 58         if(b == -1) return ;
 59         if(a == -1 || x_plus_y[a] > x_plus_y[b])
 60             a = b;
 61     }
 62 
 63     void initBit() {
 64         memset(tree + 1, -1, hcnt * sizeof(int));
 65     }
 66 
 67     void modify(int x, int val) {
 68         while(x) {
 69             update_min(tree[x], val);
 70             x -= lowbit(x);
 71         }
 72     }
 73 
 74     int query(int x) {
 75         int res = -1;
 76         while(x <= hcnt) {
 77             update_min(res, tree[x]);
 78             x += lowbit(x);
 79         }
 80         return res;
 81     }
 82 
 83     void build_edge() {
 84         sort(p, p + n);
 85         get_x_plus_y();
 86         get_y_sub_x();
 87         initBit();
 88         for(int i = n - 1; i >= 0; --i) {
 89             int tmp = query(y_sub_x[i]);
 90             if(tmp != -1) edge[ecnt++] = Edge(p[i].id, p[tmp].id, x_plus_y[tmp] - x_plus_y[i]);
 91             modify(y_sub_x[i], i);
 92         }
 93     }
 94 
 95     int fa[MAXV], ans[MAXV];
 96 
 97     int find_set(int x) {
 98         return fa[x] == x ? x : fa[x] = find_set(fa[x]);
 99     }
100 
101     int kruskal() {
102         for(int i = 0; i < n; ++i) fa[i] = i;
103         sort(edge, edge + ecnt);
104         int acnt = 0;
105         for(int i = 0; i < ecnt; ++i) {
106             int fu = find_set(edge[i].u), fv = find_set(edge[i].v);
107             if(fu != fv) {
108                 ans[acnt++] = edge[i].cost;
109                 fa[fu] = fv;
110             }
111         }
112         reverse(ans, ans + acnt);
113         return ans[k - 1];
114     }
115 
116     void mymain() {
117         scanf("%d%d", &n, &k);
118         for(int i = 0; i < n; ++i) p[i].read(i);
119 
120         build_edge();
121         for(int i = 0; i < n; ++i) swap(p[i].x, p[i].y);
122         build_edge();
123         for(int i = 0; i < n; ++i) p[i].x = -p[i].x;
124         build_edge();
125         for(int i = 0; i < n; ++i) swap(p[i].x, p[i].y);
126         build_edge();
127 
128         printf("%d\n", kruskal());
129     }
130 }
131 
132 int main() {
133     Bilibili::mymain();
134 }
View Code

 

posted @ 2015-01-31 21:34  Oyking  阅读(265)  评论(0)  编辑  收藏  举报