uvalive 4730(并查集+区间修改单点查询)

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题意：有n个城市，编号从0到n-1，给出了这n个城市的坐标，初始没有道路连通，然后有m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(连通的城市成为一个州)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，
题解：因为州是连通的城市，所以要用并查集得到连通分量，两棵线段树分别记录对应纵坐标区间内的州的数量和城市的数量，用两个数组分别记录每个连通分量的纵坐标上下界，每次有道路加入时都要更新，区间修改用模拟分类讨论。注意C是小数，为了查询对应区间，在区间修改和查询时分别把下界加1和C向上取整。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 100005;int n, m, sum1[N << 2], sum2[N << 2], addv1[N << 2], addv2[N << 2];int pa[N], num[N], up[N], down[N];char str[10];int get_parent(int x) {    return pa[x] == x ? x : pa[x] = get_parent(pa[x]);}void pushup(int k) {    sum1[k] = sum1[k * 2] + sum1[k * 2 + 1];    sum2[k] = sum2[k * 2] + sum2[k * 2 + 1];}void pushdown(int k) {    if (addv1[k]) {        addv1[k * 2] += addv1[k];        sum1[k * 2] += addv1[k];        addv1[k * 2 + 1] += addv1[k];        sum1[k * 2 + 1] += addv1[k];        addv1[k] = 0;    }    if (addv2[k]) {        addv2[k * 2] += addv2[k];        sum2[k * 2] += addv2[k];        addv2[k * 2 + 1] += addv2[k];        sum2[k * 2 + 1] += addv2[k];        addv2[k] = 0;    }}void modify(int k, int left, int right, int l1, int r1, int x, int flag) {    if (l1 > r1)        return;    if (l1 <= left && right <= r1) {        if (!flag) {            addv1[k] += x;            sum1[k] += x;        }        else {            addv2[k] += x;            sum2[k] += x;        }        return;    }    pushdown(k);    int mid = (left + right) / 2;    if (l1 <= mid)        modify(k * 2, left, mid, l1, r1, x, flag);     if (r1 > mid)        modify(k * 2 + 1, mid + 1, right, l1, r1, x, flag);    pushup(k);}int query(int k, int left, int right, int c) {    if (left == right)        return k;    pushdown(k);    int mid = (left + right) / 2;    if (c <= mid)        return query(k * 2, left, mid, c);    else        pushup(k);    if (c > mid)        return query(k * 2 + 1, mid + 1, right, c);    else        pushup(k);}int main() {    int t;    scanf("%d", &t);    while (t--) {        scanf("%d", &n);        memset(sum1, 0, sizeof(sum1));        memset(sum2, 0, sizeof(sum2));        memset(addv1, 0, sizeof(addv1));        memset(addv2, 0, sizeof(addv2));        int temp1, temp2, maxy = 0;        for (int i = 0; i < n; i++) {            scanf("%d%d", &temp1, &temp2);            pa[i] = i;            num[i] = 1;            up[i] = down[i] = temp2;            maxy = max(maxy, temp2);        }        maxy++;        scanf("%d", &m);        int x, y;        double c;        while (m--) {            scanf("%s", str);            if (str[0] == 'r') {                scanf("%d%d", &x, &y);                int px = get_parent(x);                int py = get_parent(y);                if (px != py) {                    if (up[px] > up[py])                        swap(px, py);                    if (down[py] > up[px]) {                        modify(1, 1, maxy, up[px] + 1, down[py], 1, 0);                        modify(1, 1, maxy, up[px] + 1, down[py], num[px] + num[py], 1);                        modify(1, 1, maxy, down[px] + 1, up[px], num[py], 1);                        modify(1, 1, maxy, down[py] + 1, up[py], num[px], 1);                    }                    else if (down[px] > down[py]) {                        modify(1, 1, maxy, down[px] + 1, up[px], -1, 0);                        modify(1, 1, maxy, down[py] + 1, down[px], num[px], 1);                        modify(1, 1, maxy, up[px] + 1, up[py], num[px], 1);                    }                    else {                        modify(1, 1, maxy, down[py] + 1, up[px], -1, 0);                        modify(1, 1, maxy, down[px] + 1, down[py], num[py], 1);                        modify(1, 1, maxy, up[px] + 1, up[py], num[px], 1);                    }                    pa[px] = py;                    num[py] += num[px];                    down[py] = min(down[py], down[px]);                    up[py] = max(up[py], up[px]);                }            }            else {                scanf("%lf", &c);                int k = query(1, 1, maxy, (int)(c + 1));                printf("%d %d\n", sum1[k], sum2[k]);            }        }    }    return 0;}

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