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):
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }