UVA1455 - Kingdom(并查集 + 线段树)
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#include #include 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;}