主题链接：点击打开链接
题意见白书P248
思路：
先把读入的y值都扩大2倍变成整数
然后离散化一下
用线段树来维护y轴 区间上每一个点的 城市数量和联通块数量。
然后用并查集维护每一个联通块及联通块的最大最小y值。还要加并查集的秩来记录每一个联通块的点数
然后就是模拟搞。
。
T^T绝杀失败题。。似乎数组开小了一点就过了。==
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string.h>
#include<algorithm>
using namespace std;
#define rank Rank
#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define Lson(x) tree[x].l
#define Rson(x) tree[x].r
#define Sum0(x) tree[x].sum[0]
#define Lazy0(x) tree[x].lazy[0]
#define Sum1(x) tree[x].sum[1]
#define Lazy1(x) tree[x].lazy[1]
inline int Mid(int x, int y){return (x+y)>>1;}
#define N 100005
struct Point{
int x, y;
}p[100005];
struct que{
int u, v, op;
}Q[200005];
vector<int>G;
int n, q;
char s[10];
struct node{
int l, r, sum[2], lazy[2];
}tree[N<<4];
void push_down(int id){
if(Lazy1(id)) {
Sum1(L(id)) += Lazy1(id);
Sum1(R(id)) += Lazy1(id);
Lazy1(L(id)) += Lazy1(id);
Lazy1(R(id)) += Lazy1(id);
Lazy1(id) = 0;
}
if(Lazy0(id)) {
Sum0(L(id)) += Lazy0(id);
Sum0(R(id)) += Lazy0(id);
Lazy0(L(id)) += Lazy0(id);
Lazy0(R(id)) += Lazy0(id);
Lazy0(id) = 0;
}
}
void push_up(int id){Sum0(id) = Sum0(L(id)) + Sum0(R(id));Sum1(id) = Sum1(L(id)) + Sum1(R(id));}
void build(int l, int r, int id){
Lson(id) = l; Rson(id) = r;
Sum0(id) = Lazy0(id) = Sum1(id) = Lazy1(id) = 0;
if(l == r) return ;
int mid = Mid(l, r);
build(l, mid, L(id));
build(mid+1, r, R(id));
}
void updata(int l, int r, int val, int now, int id){
push_down(id);
if(l == Lson(id) && Rson(id) == r) {
if(now==0)Sum0(id) += val, Lazy0(id) += val;
else Sum1(id) += val, Lazy1(id) += val;
return ;
}
int mid = Mid(Lson(id), Rson(id));
if(mid < l)
updata(l, r, val, now, R(id));
else if(r <= mid)
updata(l, r, val, now, L(id));
else {
updata(l, mid, val, now, L(id));
updata(mid+1, r, val, now, R(id));
}
push_up(id);
}
int query(int pos, int now, int id){
push_down(id);
if(Lson(id)==Rson(id))if(now==0)return Sum0(id); else return Sum1(id);
int mid = Mid(Lson(id), Rson(id));
int ans;
if(mid < pos)
return query(pos, now, R(id));
else return query(pos, now, L(id));
}
int f[100005], rank[100005], S[100005], X[100005]; //每一个集合的上下界
int find(int x){return x==f[x]?x:f[x] = find(f[x]);}
void Union(int x, int y){
int fx = find(x), fy = find(y);
if(fx == fy)return;
if(S[fx] > S[fy]) swap(fx, fy);
if(S[fx] <= X[fy]){
updata(S[fx], X[fy], 1, 0, 1);
updata(S[fx], X[fy], rank[fx] + rank[fy], 1, 1);
updata(X[fx], S[fx], rank[fy], 1, 1);
updata(X[fy], S[fy], rank[fx], 1, 1);
}
else if(X[fx] >= X[fy]) {
updata(X[fx], S[fx], -1, 0, 1);
updata(X[fy], X[fx], rank[fx], 1, 1);
updata(S[fx], S[fy], rank[fx], 1, 1);
}
else {
updata(X[fy], S[fx], -1, 0, 1);
updata(X[fx], X[fy], rank[fy], 1, 1);
updata(S[fx], S[fy], rank[fx], 1, 1);
}
if(rank[fy]<rank[fx])swap(fx, fy);
f[fx] = fy;
rank[fy] += rank[fx];
rank[fx] = 0;
X[fy] = min(X[fy], X[fx]);
S[fy] = max(S[fy], S[fx]);
}
void input(){
G.clear();
scanf("%d", &n);
for(int i = 1; i <= n; i++) f[i] = i, rank[i] = 1;
for(int i = 1; i <= n; i++) scanf("%d %d",&p[i].x, &p[i].y), p[i].y <<= 1, G.push_back(p[i].y);
scanf("%d", &q);
for(int i = 0; i < q; i++)
{
scanf("%s", s);
if(s[0]=='r')
{
Q[i].op = 1;
scanf("%d %d", &Q[i].u, &Q[i].v); Q[i].u++; Q[i].v++;
}
else {
Q[i].op = 2;
scanf("%d.5",&Q[i].u);
Q[i].u = Q[i].u * 2+1;
G.push_back(Q[i].u);
}
}
sort(G.begin(), G.end());
G.erase(unique(G.begin(), G.end()), G.end());
for(int i = 1; i <= n; i++)X[i] = S[i] = p[i].y = lower_bound(G.begin(), G.end(), p[i].y) - G.begin()+1;
for(int i = 0; i < q; i++)if(Q[i].op == 2)Q[i].u = lower_bound(G.begin(), G.end(), Q[i].u) - G.begin()+1;
}
int main() {
int T; scanf("%d",&T);
while(T--){
input();
build(1, G.size(), 1);
for(int i = 0; i < q; i++) {
if(Q[i].op == 1)
{
Union(Q[i].u, Q[i].v);
}
else
printf("%d %d
", query(Q[i].u, 0, 1), query(Q[i].u, 1, 1));
}
}
return 0;
}
/*
3
11
1 7
5 7
8 6
3 5
5 5
2 3
10 3
7 2
4 1
11 1
4 6
21
road 0 1
road 3 5
line 6.5
road 4 2
road 3 8
road 4 7
road 6 9
road 4 1
road 2 7
line 4.5
line 6.5
line 3.5
line 2.5
line 5.5
road 10 0
line 5.5
line 6.5
road 0 3
line 1.5
line 6.5
line 2.5
ans:
0 0
2 8
1 5
2 8
3 10
1 5
1 6
1 6
2 11
1 9
2 11
*/版权声明：本文博客原创文章，博客，未经同意，不得转载。