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http://poj.org/problem?id=3241
曼哈顿距离最小生成树模板题。
核心思想是把坐标系转3次,以及以横坐标为第一关键字,纵坐标为第二关键字排序后,从后往前扫。扫完一个点就把它插到树状数组的x-y位置上,权值为x+y。查询时查询扫过的所有点满足xdone-ydone>=xnow-xnow时,直接是树状数组中的的一个后缀区间,从后往前扫保证了区间内的这些点都在当前点的y轴向右扫45度的范围内。树状数组实现查询x+y的最小值,以及此最小值对应原数组中的位置,方便建图连边。
模板是抄的别人的QAQ
时间复杂度$O(nlogn)$
#include<cmath> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 100003; int in() { int k = 0, fh = 1; char c = getchar(); for(; c < ‘0‘ || c > ‘9‘; c = getchar()) if (c == ‘-‘) fh = -1; for(; c >= ‘0‘ && c <= ‘9‘; c = getchar()) k = (k << 3) + (k << 1) + c - ‘0‘; return k * fh; } struct Point { int x, y, id; bool operator < (const Point &A) const { return x == A.x ? y < A.y : x < A.x; } } P[N]; struct Bits { int mn, pos; void init() {mn = 0x7fffffff; pos = -1;} } bit[N]; int tot; struct Edge { int u, v, dis; bool operator < (const Edge &A) const { return dis < A.dis; } } E[N << 2]; void add(int a, int b, int c) {E[++tot] = (Edge) {a, b, c};} int n, fa[N], k; int find(int x) { if (fa[x] == x) return x; fa[x] = find(fa[x]); return fa[x]; } int dist(int x, int y) { return abs(P[x].x - P[y].x) + abs(P[x].y - P[y].y); } void update(int x, int num, int pos) { for(; x; x -= (x & (-x))) if (num < bit[x].mn) bit[x].mn = num, bit[x].pos = pos; } int m; int query(int x) { int ans = 0x7fffffff, pos = -1; for(x; x <= m; x += (x & (-x))) if (bit[x].mn < ans) ans = bit[x].mn, pos = bit[x].pos; return pos; } int a[N], H[N], cnt; int solve() { for(int change = 0; change < 4; ++change) { if (change == 1 || change == 3) for(int i = 1; i <= n; ++i) swap(P[i].x, P[i].y); else if (change == 2) for(int i = 1; i <= n; ++i) P[i].x = -P[i].x; sort(P + 1, P + n + 1); for(int i = 1; i <= n; ++i) a[i] = H[i] = P[i].y - P[i].x; cnt = n; sort(H + 1, H + cnt + 1); cnt = unique(H + 1, H + cnt + 1) - H; for(int i = 1; i <= cnt; ++i) bit[i].init(); int pos, tmp; m = cnt; for(int i = n; i > 0; --i) { pos = lower_bound(H + 1, H + cnt, a[i]) - H; tmp = query(pos); if (tmp != -1) add(P[i].id, P[tmp].id, dist(i, tmp)); update(pos, P[i].x + P[i].y, i); } } sort(E + 1, E + tot + 1); cnt = n - k; for(int i = 1; i <= n; ++i) fa[i] = i; int u, v; for(int i = 1; i <= tot; ++i) { u = find(E[i].u); v = find(E[i].v); if (u != v) { --cnt; fa[u] = v; if (cnt == 0) return E[i].dis; } } } int main() { while (~scanf("%d%d", &n, &k)) { tot = 0; for(int i = 1; i <= n; ++i) { P[i].x = in(); P[i].y = in(); P[i].id = i; } printf("%d\n", solve()); } return 0; }
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原文地址:http://www.cnblogs.com/abclzr/p/5796951.html