标签:style http color io os ar for sp on
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,还有每个州的y的上下界。建立坐标y的线段树,然后每次执行road操作的时候,对范围内的y坐标进行更新;更新需要分三种情况:两个州是相离,还是相交,还是包含(这里指的是y坐标的关系);并且因为这里查询是浮点数,所以更新的时候[l,r]的时候,只更新[l,r),这里的r留给它下面的点,这样每次查询的时候就可以查询(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]) {//包含
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;
}
标签:style http color io os ar for sp on
原文地址:http://blog.csdn.net/u012997373/article/details/40342167