题目链接：uva 1455 - Kingdom

题目大意：平面上又n个城市，初始时城市之间没有任何双向道路相连，要求一次执行指令。

• road A B ：在城市A和城市B之间连接一条双向道路
• line C：询问一条y=C的水平线上穿过多少州和这些州总共有多少城市。
  一个联通分量算一个州，C保证为小数部分为0.5的实数。

解题思路：线段树维护每个位置上州和城市的个数，并查集维护哪些城市属于同一个州，并且要记录这些州上下范围。每次新建一条道路，要相应根据两个州的y坐标范围对线段树进行维护。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1e6;
const int maxp = 1e5+5;
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)+1)

struct Node {
    int l, r, nc, nz;
    void set (int l, int r, int nc, int nz) {
        this->l = l;
        this->r = r;
        this->nc = nc;
        this->nz = nz;
    }
}node[maxn * 4 + 5];

struct Segment {
    int l, r;
    Segment (int l = 0, int r = 0) {
        this->l = l;
        this->r = r;
    }
}s[maxp];

void add_node(int u, int vc, int vz) {
    /*
    if (node[u].l == node[u].r && node[u].l == 1)
        printf("solve before:%d %d %d %d!\n", node[u].nc, node[u].nz, vc, vz);
        */
    node[u].nc += vc;
    node[u].nz += vz;
}

void pushup(int u) {
    node[u].set(node[lson(u)].l, node[rson(u)].r, 0, 0);
}

void pushdown(int u) {
    add_node(lson(u), node[u].nc, node[u].nz);
    add_node(rson(u), node[u].nc, node[u].nz);
    node[u].nc = node[u].nz = 0;
}

void build_segTree (int u, int l, int r) {
    if (l == r) {
        node[u].set(l, r, 0, 0);
        return;
    }

    int mid = (l + r) / 2;
    build_segTree(lson(u), l, mid);
    build_segTree(rson(u), mid + 1, r);
    pushup(u);
}

void insert_segTree (int u, int l, int r, int vc, int vz) {
    if (l <= node[u].l && node[u].r <= r) {
        add_node(u, vc, vz);
        return;
    }

    pushdown(u);
    int mid = (node[u].l + node[u].r) / 2;
    if (l <= mid)
        insert_segTree(lson(u), l, r, vc, vz);
    if (r > mid)
        insert_segTree(rson(u), l, r, vc, vz);
}

Segment query_segTree (int u, int x) {
    if (node[u].l == x && node[u].r == x)
        return Segment(node[u].nc, node[u].nz);

    pushdown(u);
    int mid = (node[u].l + node[u].r) / 2;
    if (x <= mid)
        return query_segTree(lson(u), x);
    else
        return query_segTree(rson(u), x);
}

int f[maxp], d[maxp];

void init () {
    build_segTree (1, 0, maxn);

    int n, x, y;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        f[i] = i;
        d[i] = 1;
        scanf("%d%d", &x, &y);
        s[i].l = s[i].r = y;
    }
}

void change (int r, int l, int vc, int vz) {
    if (r < l)
        return;
    insert_segTree (1, l, r, vc, vz);
}

int getfar (int x) {
    return x == f[x] ? x : f[x] = getfar(f[x]);
}

void link (int u, int v) {
    int x = getfar(u);
    int y = getfar(v);

    if (x == y)
        return;

    if (s[x].l > s[y].l)
        swap(x, y);

    //printf("%d %d %d %d!!!\n", s[x].l, s[x].r, s[y].l, s[y].r);

    int root = 1048576;

    if (s[y].l >= s[x].r) {
        change(s[y].l - 1, s[x].r, 0, 1);
        change(s[y].r - 1, s[x].r, d[x], 0);
    } else if (s[y].r <= s[x].r) {
        change(s[y].r - 1, s[y].l, 0, -1);
        change(s[x].r - 1, s[y].r, d[y], 0);
    } else {
        change(s[x].r - 1, s[y].l, 0, -1);
        change(s[y].r - 1, s[x].r, d[x], 0);
    }
    //printf("fuck:%d %d\n", node[root].nc, node[root].nz);
    change(s[y].l - 1, s[x].l, d[y], 0);


    s[y].l = min(s[x].l, s[y].l);
    s[y].r = max(s[x].r, s[y].r);
    f[x] = y;
    d[y] += d[x];
}

void solve () {
    int n, x, y;
    double q;
    char order[10];

    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%s", order);
        if (strcmp(order, "road") == 0) {
            scanf("%d%d", &x, &y);
            link(x, y);
        } else {
            scanf("%lf", &q);
            Segment ans = query_segTree(1, (int)q);
            printf("%d %d\n", ans.r, ans.l);
        }
    }
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        init();
        solve();
    }
    return 0;
}

uva 1455 - Kingdom(并查集+线段树)