UVA1455 - Kingdom(并查集 + 线段树)

题目链接

题目大意：一个平面内，给你n个整数点，两种类型的操作：road x y 把city x 和city y连接起来，line fnum （浮点数小数点一定是0.5） 查询y = fnum这条直线穿过了多少个州和city。州指的是连通的城市。

解题思路：用并查集记录城市之间是否连通，还有每一个州的y的上下界。建立坐标y的线段树，然后每次运行road操作的时候，对范围内的y坐标进行更新；更新须要分三种情况：两个州是相离，还是相交，还是包括（这里指的是y坐标的关系）；而且由于这里查询是浮点数，所以更新的时候[l,r]的时候，仅仅更新[l,r)，这里的r留给它以下的点，这样每次查询的时候就能够查询（int）fnum。

代码：

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

using namespace std;

const int maxn = 1e6 + 5;
const int N = 1e5 + 5;
#define lson(x) (x<<1)
#define rson(x) ((x<<1) | 1)

struct Node {

    int l, r, ns, nc;
    void set (int l, int r, int ns, int nc) {

        this->l = l;
        this->r = r;
        this->ns = ns;
        this->nc = nc;
    }
}node[4 * maxn];

void pushup (int u) {

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

void add_node (int u, int adds, int addc) {

    node[u].ns += adds;
    node[u].nc += addc;    
}

void pushdown (int u) {

    if (node[u].ns || node[u].nc) {
        add_node(lson(u), node[u].ns, node[u].nc);
        add_node(rson(u), node[u].ns, node[u].nc);    
    }
}

void build (int u, int l, int r) {

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

    int m = (l + r)>>1;
    build (lson(u), l, m);
    build (rson(u), m + 1, r);
    pushup(u);
}

void update (int u, int l, int r, int addc, int adds) {

    if (node[u].l >= l && node[u].r <= r) {
        add_node (u, adds, addc);    
        return;
    }

    int m = (node[u].l + node[u].r)>>1;
    pushdown(u);
    if (l <= m)
        update (lson(u), l, r, addc, adds);
    if (r > m)
        update (rson(u), l, r, addc, adds);
    pushup(u);
}

int query (int u, int x) {

    if (node[u].l == x && node[u].r == x)
        return u;

    int m = (node[u].l + node[u].r)>>1;
    int ans;

    pushdown(u);
    if (x <= m)
        ans = query (lson(u), x);
    else
        ans = query (rson(u), x);
    pushup(u);
    return ans;
}

int p[N], cnt[N], L[N], R[N];
int n, m;

int getParent (int x) {
    return x == p[x] ? x: p[x] = getParent (p[x]);
}

void change (int u, int l, int r, int addc, int adds) {

    if (r < l) //注意
        return;
    update (u, l, r, addc, adds); 
}

void Union(int x, int y) {

    x = getParent (x);
    y = getParent (y);

    if (x == y)
        return;

    if (L[x] >= L[y])
        swap(x, y);

    if (R[x] <= L[y]) {//相离

        change (1, L[x], R[x] - 1, cnt[y], 0);        
        change (1, L[y], R[y] - 1, cnt[x], 0);
        change (1, R[x], L[y] - 1, cnt[x] + cnt[y], 1);
    } else if (R[y] <= R[x]) {//包括

        change (1, L[x], L[y] - 1, cnt[y], 0);
        change (1, R[y], R[x] - 1, cnt[y], 0);
        change (1, L[y], R[y] - 1, 0, -1);
    } else {//相交

        change (1, L[x], L[y] - 1, cnt[y], 0);
        change (1, R[x], R[y] - 1, cnt[x], 0);
        change (1, L[y], R[x] - 1, 0, -1);
    }

    p[x] = y;
    cnt[y] += cnt[x];
    L[y] = min (L[y], L[x]);
    R[y] = max (R[y], R[x]);
}

void init () {

    int x, y;
    scanf ("%d", &n);

    for (int i = 0; i < n; i++) {
        scanf ("%d%d", &x, &y);
        p[i] = i;
        cnt[i] = 1;
        L[i] = R[i] = y;

    }

    scanf ("%d", &m);
    build (1, 0, maxn - 5);
}

void solve () {

    char str[100];
    int x, y;
    double q;

    for (int i = 0; i < m; i++) {

        scanf ("%s", str);

        if (str[0] == ‘r‘) {
            scanf ("%d%d", &x, &y);        
            Union(x, y);
        } else {
            scanf ("%lf", &q);
            x = query (1, (int)q);
            printf ("%d %d\n", node[x].ns, node[x].nc);
        }
    }
}

int main () {

    int T;
    scanf ("%d", &T);

    while (T--) {

        init();
        solve();    
    }
    return 0;
}