/*
APIO 2017 Rainbow
- We can view a rectangle as a planar graph
- Put temporary rivers on the outside of the rectangle
- Each river segment is a node and adjacent rivers constitute an edge
- This problem then becomes finding the number of faces of a planar graph
- We can solve this with Euler's formula
- By Euler's formula, we have
F = E - V + 1 + C
where F is faces, E is edges, V is vertices, and C is the number of components
- Each corner of a square is a vertex and each side of a square is an edge if it has a river
*/
#include "rainbow.h"
#include <bits/stdc++.h>
#define FOR(i, x, y) for (int i = x; i < y; i++)
using namespace std;
const int MAXN = 2e5, MAXSEG = (6e5 + 9) * 19 + 1;
int cnt = 1, segtree[MAXSEG], left_c[MAXSEG], right_c[MAXSEG];
struct Segtree {
set<int> data[MAXN + 1];
int roots[MAXN + 2];
void add(int x, int y) { data[x].insert(y); }
void build() {
FOR(i, 1, MAXN + 1) {
roots[i + 1] = roots[i];
for (int j : data[i]) update(j, roots[i + 1]);
}
}
void update(int pos, int &node, int l = 1, int r = MAXN) {
segtree[cnt] = segtree[node] + 1;
left_c[cnt] = left_c[node];
right_c[cnt] = right_c[node];
node = cnt++;
if (l == r) return;
int mid = (l + r) / 2;
if (pos > mid) update(pos, right_c[node], mid + 1, r);
else update(pos, left_c[node], l, mid);
}
int query(int l1, int r1, int l2, int r2) {
if (l2 > r2) return 0;
return query(l2, r2, roots[r1 + 1], 1, MAXN) - query(l2, r2, roots[l1], 1, MAXN);
}
int query(int a, int b, int node, int l, int r) {
if (a > r || b < l) return 0;
if (a <= l && b >= r) return segtree[node];
int mid = (l + r) / 2;
return query(a, b, left_c[node], l, mid) + query(a, b, right_c[node], mid + 1, r);
}
} vertices, edges_horiz, edges_vert, rivers;
int mx_r, mn_r, mx_c, mn_c;
void add_river(int x, int y) {
vertices.add(x, y);
vertices.add(x + 1, y);
vertices.add(x, y + 1);
vertices.add(x + 1, y + 1);
edges_horiz.add(x, y);
edges_horiz.add(x + 1, y);
edges_vert.add(x, y);
edges_vert.add(x, y + 1);
rivers.add(x, y);
}
void init(int R, int C, int sr, int sc, int M, char *S) {
add_river(sr, sc);
mx_r = mn_r = sr;
mx_c = mn_c = sc;
FOR(i, 0, M) {
if (S[i] == 'N') sr--;
if (S[i] == 'E') sc++;
if (S[i] == 'S') sr++;
if (S[i] == 'W') sc--;
add_river(sr, sc);
mx_r = max(mx_r, sr);
mn_r = min(mn_r, sr);
mx_c = max(mx_c, sc);
mn_c = min(mn_c, sc);
}
vertices.build();
edges_horiz.build();
edges_vert.build();
rivers.build();
}
int colour(int ar, int ac, int br, int bc) {
int E = edges_horiz.query(ar + 1, br, ac, bc) + edges_vert.query(ar, br, ac + 1, bc);
int V = vertices.query(ar + 1, br, ac + 1, bc);
int R = rivers.query(ar, br, ac, bc);
int C = (ar >= mn_r || br <= mx_r || ac >= mn_c || bc <= mx_c ? 1 : 2);
return E - V + C - R;
}
