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);elseans = 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;
}