UVA1455 - Kingdom(并查集 + 线段树)

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
, cnt
, L
, R
;
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;
}