Uva4730 Kingdom (并查集+线段树)

题意：有n个城市，给出这n个城市的坐标，初始没有道路连通，有q个指令，

road A B：在城市A和城市B之间连接一条双向道路，保证这条道路不和其他道路在非端点处相交。

line C：询问y = C的水平线和多少个州相交，以及这些州一共包含多少座城市，C小数部分保证为0.5

思路：记录每个洲的纵坐标最小值最大值， 当连接A, B时， 如果A, B处于同一连通块，那么不做处理， 否则， 在A所在连通块的纵坐标范围内州数减去1，城市数减去所在连通块的城市数，B也是如此， 然后连接起来之后的连通块纵坐标范围内加上原来减去的和， 州数加1， lazy标记一下即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<iostream>
#include<algorithm>
const int maxn = 1e5 + 10;
const int INF = 1e9;
using namespace std;

struct msg {
    int city, sta;
} C[maxn * 20];
int n, x, T, q, A, B;
int ymax[maxn], ymin[maxn];
int d[maxn * 4], y[maxn];
int sum[maxn];
int pa[maxn], int_line;
char op[10];

void init() {
    for(int i = 0; i < maxn; i++) {
        sum[i] = 1; pa[i] = i;
    }

    for(int i = 0; i < 20 * maxn; i++) {
        C[i].city = 0;
        C[i].sta = 0;
    }
}

int findset(int x) { return x == pa[x] ? x : pa[x] = findset(pa[x]); }

int ql, qr, city, sta;
void pushdown(int o) {
    int o1 = o << 1, o2 = o << 1 | 1;
    C[o1].city += C[o].city;
    C[o2].city += C[o].city;
    C[o].city = 0;
    C[o1].sta += C[o].sta;
    C[o2].sta += C[o].sta;
    C[o].sta = 0;
}

void update(int o, int l, int r) {
    if(l > qr || r < ql) return ;
    if(l >= ql && r <= qr) {
        C[o].city += city;
        C[o].sta += sta;
        return ;
    }
    pushdown(o);
    int mid = (l + r) >> 1;
    int o1 = o << 1, o2 = o << 1 | 1;
    if(ql <= mid) update(o1, l, mid);
    if(qr > mid) update(o2, mid + 1, r);
}

msg query(int o, int l, int r) {
    if(l == r) return C[o];
    int mid = (l + r) >> 1;
    int o1 = o << 1, o2 = o << 1 | 1;
    pushdown(o);
    if(x <= mid) return query(o1, l, mid);
    else return query(o2, mid + 1, r);
}

void unit(int a, int b, int R) {
    int na = findset(a);
    int nb = findset(b);
    if(na == nb) return ;
    int nax = ymin[na], nay = ymax[na];
    int nbx = ymin[nb], nby = ymax[nb];
    sta = -1; city = -sum[na];
    ql = lower_bound(d, d + R, nax) - d;
    qr = lower_bound(d, d + R, nay) - d;
    update(1, 0, R);

    sta = -1; city = -sum[nb];
    ql = lower_bound(d, d + R, nbx) - d;
    qr = lower_bound(d, d + R, nby) - d;
    update(1, 0, R);

    pa[na] = nb; sum[nb] += sum[na];
    ymin[nb] = min(ymin[na], ymin[nb]);
    ymax[nb] = max(ymax[na], ymax[nb]);
    sta = 1; city = sum[nb];
    ql = lower_bound(d, d + R, ymin[nb]) - d;
    qr = lower_bound(d, d + R, ymax[nb]) - d;
    update(1, 0, R);
}

int main() {
    scanf("%d", &T);
    while(T--) {
        scanf("%d", &n);
        init();
        int nn = 0;
        for(int i = 0; i < n; i++) {
            scanf("%d %d", &x, &y[i]);
            y[i] *= 2;
            ymax[i] = ymin[i] = y[i];
            d[nn++] = y[i];
            d[nn++] = y[i] + 1;
            d[nn++] = y[i] - 1;
        }
        sort(d, d + nn);
        nn = unique(d, d + nn) - d;
        for(int i = 0; i < n; i++) {
            x = lower_bound(d, d + nn, y[i]) - d;
            ql = qr = x;
            city = sta = 1;
            update(1, 0, nn);
        }

        scanf("%d", &q);
        while(q--) {
            scanf("%s", op);
            if(op[0] == 'r') {
                scanf("%d %d", &A, &B);
                unit(A, B, nn);
            } else {
                double c;
                scanf("%lf", &c);
                int_line = (int)(c * 2);
                x = lower_bound(d, d + nn, int_line) - d;
                msg ans = query(1, 0, nn);
                printf("%d %d\n", ans.sta, ans.city);
            }
        }
    }
    return 0;
}