并查集 + 线段树 LA 4730 Kingdom

Posted 2020-06-17

 

题目传送门

题意：训练指南P248

分析：第一个操作可以用并查集实现，保存某集合的最小高度和最大高度以及城市个数。运用线段树成端更新来统计一个区间高度的个数，此时高度需要离散化。这题两种数据结构一起使用，联系紧密。

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 5;
const int M = 3 * N;
const int INF = 0x3f3f3f3f;
struct Point    {
    int x, y;
    Point() {}
    Point(int x, int y) : x (x), y (y) {}
};
struct Query    {
    int op;
    int u, v, w;
}q[2*N];

#define lson l, mid, o << 1
#define rson mid + 1, r, o << 1 | 1
int city[M<<2], num[M<<2], col[M<<2], col2[M<<2];          //Segment_Tree
void push_down(int o)   {
    if (col[o])   {
        col[o<<1] += col[o];    col[o<<1|1] += col[o];
        city[o<<1] += col[o];  city[o<<1|1] += col[o];
        col[o] = 0;
    }
    if (col2[o])    {
        col2[o<<1] += col2[o];   col2[o<<1|1] += col2[o];
        num[o<<1] += col2[o];  num[o<<1|1] += col2[o];
        col2[o] = 0;
    }
}
void build(int l, int r, int o) {
    col[o] = 0;    col2[o] = 0;
    if (l == r) {
        city[o] = num[o] = 0;
        return ;
    }
    int mid = l + r >> 1;
    build (lson);   build (rson);
}
void updata(int ql, int qr, int c1, int c2, int l, int r, int o)    {
    if (ql <= l && r <= qr) {
        col[o] += c1;    col2[o] += c2;
        city[o] += c1;    num[o] += c2;
        return ;
    }
    push_down (o);
    int mid = l + r >> 1;
    if (ql <= mid)  updata (ql, qr, c1, c2, lson);
    if (qr > mid)   updata (ql, qr, c1, c2, rson);
}
void query(int p, int l, int r, int o)  {
    if (l == r && l == p)   {
        printf ("%d %d\n", city[o], num[o]);   return ;
    }
    push_down (o);
    int mid = l + r >> 1;
    if (p <= mid)   query (p, lson);
    else    query (p, rson);
}

int n, m, bound;

int rt[N], rk[N], maxy[N], miny[N];             //DSU
void init(void) {
    memset (rt, -1, sizeof (rt));
    memset (rk, 0, sizeof (rk));
}
int Find(int x) {
    return rt[x] == -1 ? x : rt[x] = Find (rt[x]);
}
void Union(int u, int v)    {
    u = Find (u);   v = Find (v); 
    if (u == v) return ;
    if (rk[u] > rk[v])  swap (u, v);
    if (rk[u])  updata (miny[u], maxy[u], -1, -(rk[u] + 1), 1, bound, 1);
    if (rk[v])  updata (miny[v], maxy[v], -1, -(rk[v] + 1), 1, bound, 1);
    
    maxy[v] = max (maxy[v], maxy[u]);
    miny[v] = min (miny[v], miny[u]);
    
    rt[u] = v;  rk[v] += rk[u] + 1;
    rk[u] = 0;
    updata (miny[v], maxy[v], 1, rk[v] + 1, 1, bound, 1);
}

Point point[N];
vector<int> ys;

void run(void)  {
    init ();    build (1, bound, 1);
    for (int i=1; i<=n; ++i)    {
        miny[i] = maxy[i] = point[i].y;
    }
    for (int i=1; i<=m; ++i)    {
        if (q[i].op == 0)   {
            Union (q[i].u, q[i].v);
        }
        else    {
            query (q[i].w, 1, bound, 1);
        }
    }
}

int main(void)  {
    int T;  scanf ("%d", &T);
    while (T--) {
        scanf ("%d", &n);
        int x, y;
        ys.clear ();
        for (int i=1; i<=n; ++i) {
            scanf ("%d%d", &x, &y); point[i] = Point (x, 2 * y);
            ys.push_back (point[i].y);
        }
        char str[10];
        int u, v;    double t;
        scanf ("%d", &m);
        for (int i=1; i<=m; ++i)    {
            scanf ("%s", &str);
            if (str[0] == ‘r‘)  {
                scanf ("%d%d", &u, &v);
                q[i].op = 0;    q[i].u = u + 1, q[i].v = v + 1;
            }
            else if (str[0] == ‘l‘) {
                scanf ("%lf", &t);
                q[i].op = 1;    q[i].w = (int) (2 * t);
                ys.push_back (q[i].w);
            }
        }
        sort (ys.begin (), ys.end ());
        ys.erase (unique (ys.begin (), ys.end ()), ys.end ());
        bound = 200010;
        for (int i=1; i<=n; ++i)    {
            point[i].y = lower_bound (ys.begin (), ys.end (), point[i].y) - ys.begin () + 1;
        }
        for (int i=1; i<=m; ++i)    {
            if (q[i].op == 1)   {
                q[i].w = lower_bound (ys.begin (), ys.end (), q[i].w) - ys.begin () + 1;
            }
        }
        run ();
    }

    return 0;
}