BZOJ 2177 [曼哈顿最小生成树]

来源：互联网发布：mac 强制清理垃圾桶编辑：程序博客网时间：2024/04/30 20:13

Description

平面坐标系xOy内，给定n个顶点V=(x,y)。对于顶点u,v,u与v之间的距离d定义为|xu–xv|+|yu–yv| 。你的任务就是求出这n个顶点的最小生成树。

Solution

把平面划分为八个区域以后只有这八个区域的最近点与该点的连边在Kruscal中有贡献。
找到这八个点只要用树状数组维护一下即可。

好像很妙的方法。
有一个地方要去重，不然会WA

#include <cstdio>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;const int N = 101010;const int INF = 2147483647;inline char get(void) {    static char buf[100000], *S = buf, *T = buf;    if (S == T) {        T = (S = buf) + fread(buf, 1, 100000, stdin);        if (S == T) return EOF;    }    return *S++;}inline void read(int &x) {    static char c; x = 0; int sgn = 0;    for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;    for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';    if (sgn) x = -x;}struct Point {    int x, y, id;    Point (int _x = 0, int _y = 0, int i = 0):x(_x), y(_y), id(i) {}    inline friend bool operator <(const Point &a, const Point &b) {        return a.x == b.x ? a.y < b.y : a.x < b.x;    }    inline void Ref(int dir) {        if (dir & 1) swap(x, y);        else if (dir == 2) x = -x;    }};struct node {    int pos, key;    node (int p = 0, int k = 0):pos(p), key(k) {}};struct edge {    int from, to, key;    edge(int f = 0, int t = 0, int k = 0):from(f), to(t), key(k) {}    inline friend bool operator <(const edge &a, const edge &b) {        return a.key < b.key;    }};node C[N];Point P[N];edge G[N << 3];int n, pos, Gcnt, cnt;long long ans;int mp[N], v[N];int fa[N], rk[N];inline int lowbit(int x) {    return x & -x;}inline void Modify(int x, int key, int pos) {    for (; x; x -= lowbit(x))        if (C[x].key > key)            C[x] = node(pos, key);}inline int Query(int x) {    int key = INF, pos = -1;    for (; x <= n; x += lowbit(x))        if (C[x].key < key) {            key = C[x].key; pos = C[x].pos;        }    return pos;}inline int Abs(int x) {    return x < 0 ? -x : x;}inline int Dis(const Point &a, const Point &b) {    return Abs(a.x - b.x) + Abs(a.y - b.y);}inline int F(int x) {    return fa[x] == x ? x : fa[x] = F(fa[x]);}inline bool Merge(int x, int y) {    static int f1, f2;    f1 = F(x); f2 = F(y);    if (f1 == f2) return false;    if (rk[f1] > rk[f2]) swap(f1, f2);    if (rk[f1] == rk[f2]) rk[f2]++;    fa[f1] = f2; return true;}inline bool cmp(const int a, const int b) {    return v[a] < v[b];}inline void AddEdge(int from, int to, int key) {    G[++Gcnt] = edge(from, to, key);}int main(void) {    freopen("1.in", "r", stdin);    freopen("1.out", "w", stdout);    read(n);    for (int i = 1; i <= n; i++) {        read(P[i].x); read(P[i].y);        P[i].id = i; fa[i] = i;    }    for (int dir = 0; dir < 4; dir++) {        for (int i = 1; i <= n; i++) P[i].Ref(dir);        sort(P + 1, P + n + 1);        for (int i = 1; i <= n; i++) {            mp[i] = i; v[i] = P[i].y - P[i].x;            C[i] = node(-1, INF);        }        sort(mp + 1, mp + n + 1, cmp);        cnt = 0; mp[n + 1] = INF;        for (int i = 1; i <= n; i++) {            ++cnt;            while (v[mp[i]] == v[mp[i + 1]]) v[mp[i++]] = cnt;            v[mp[i]] = cnt;        } // 离散去重        for (int i = n; i; i--) {            pos = Query(v[i]);            if (~pos) AddEdge(P[i].id, P[pos].id, Dis(P[i], P[pos]));            Modify(v[i], P[i].x + P[i].y, i);        }    }    sort(G + 1, G + Gcnt + 1);    for (int i = 1; i <= Gcnt; i++) {        if (Merge(G[i].from, G[i].to)) {            n--; ans += G[i].key;            if (n == 1) break;        }    }    cout << ans << endl;}

0 0