Submission #885416

#	Time	Username	Problem	Language	Result	Execution time	Memory
885416		pubin06	Land of the Rainbow Gold (APIO17_rainbow)	C++14	12 / 100	165 ms	193384 KiB

rainbow

// #include "rainbow.h"
#include <bits/stdc++.h>
#define sz(v) (int)(v).size()
using namespace std;
 
const int MXn = 2e5 + 5;
const int dx[] = {1, 0, 1};
const int dy[] = {0, 1, 1};
 
int n, m;
struct PST {
	vector<int> val, idL, idR;
	void resiz(int s) {
		val.resize(s + 3, 0);
		idL.resize(s + 3, -1);
		idR.resize(s + 3, -1);
	}
	int no = 0;
	int build(int l = 1, int r = n) {
		int cur = ++no;
		assert(cur < sz(val));
		val[cur] = 0;
		if (l == r) return cur;
		int mid = (l + r) >> 1;
		idL[cur] = build(l, mid);
		idR[cur] = build(mid + 1, r);
		return cur;
	}
	int update(int i, int v, int oID, int l = 1, int r = n) {
		int cur = ++no;
		assert(cur < sz(val));
		if (l == r) {
			val[cur] = val[oID] + v;
			return cur;
		}
		int mid = (l + r) >> 1;
		if (i <= mid) {
			idL[cur] = update(i, v, idL[oID], l, mid);
			idR[cur] = idR[oID];
		} else {
			idL[cur] = idL[oID];
			idR[cur] = update(i, v, idR[oID], mid + 1, r);
		}
		val[cur] = val[idL[cur]] + val[idR[cur]];
		return cur;
	}
	int get(int Lf, int Rt, int id, int l = 1, int r = n) {
	    assert(1 <= Lf && Rt <= n);
		if (Lf <= l && r <= Rt) {
			return val[id];
		}
		int mid = (l + r) >> 1;
		int res = 0;
		if (Lf <= mid) res += get(Lf, Rt, idL[id], l, mid);
		if (Rt > mid) res += get(Lf, Rt, idR[id], mid + 1, r);
		return res;
	}
} T[4];
int root[4][MXn];
 
bool onboard(int x, int y) {
	return 1 <= x && x <= n && 1 <= y && y <= m;
}
vector<int> river[MXn], v1[MXn], v2[MXn], v3[MXn];
void add(int sr, int sc) {
    assert(onboard(sr, sc));
	river[sc].emplace_back(sr);
	if (sc > 1) v1[sc - 1].emplace_back(sr);
	v1[sc].emplace_back(sr);
	if (sr > 1) v2[sc].emplace_back(sr - 1);
	v2[sc].emplace_back(sr);
	v3[sc].emplace_back(sr);
	for (int k = 0; k < 3; k++) {
		int xx = dx[k] + sr, yy = dy[k] + sc;
		if (onboard(xx, yy)) v3[yy].emplace_back(xx);
	}
}
void init(int R, int C, int sr, int sc, int M, char *S) {
	n = R, m = C;
	add(sr, sc);
	for (int i = 0; i < M; i++) {
		if (S[i] == 'N') sr--;
		if (S[i] == 'S') sr++;
		if (S[i] == 'W') sc--;
		if (S[i] == 'E') sc++;
		add(sr, sc);
	}
	vector<int> siz(4, 0);
	for (int c = 1; c <= C; c++) {
		sort(begin(river[c]), end(river[c]));
		river[c].resize(unique(begin(river[c]), end(river[c])) - river[c].begin());
		siz[0] += sz(river[c]);
		sort(begin(v1[c]), end(v1[c]));
		v1[c].resize(unique(begin(v1[c]), end(v1[c])) - v1[c].begin());
		siz[1] += sz(v1[c]);
		sort(begin(v2[c]), end(v2[c]));
		v2[c].resize(unique(begin(v2[c]), end(v2[c])) - v2[c].begin());
		siz[2] += sz(v2[c]);
		sort(begin(v3[c]), end(v3[c]));
		v3[c].resize(unique(begin(v3[c]), end(v3[c])) - v3[c].begin());
		siz[3] += sz(v3[c]);
	}
	for (int t = 0; t < 4; t++) {
		T[t].resiz((1 << 19) + siz[t] * 19);
		root[t][0] = T[t].build();
	}
	for (int c = 1; c <= C; c++) {
		root[0][c] = root[0][c - 1];
		root[1][c] = root[1][c - 1];
		root[2][c] = root[2][c - 1];
		root[3][c] = root[3][c - 1];
		
		for (int r: river[c]) {
			root[0][c] = T[0].update(r, 1, root[0][c]);
		}
		for (int r: v1[c]) {
			root[1][c] = T[1].update(r, 1, root[1][c]);
		}
		for (int r: v2[c]) {
			root[2][c] = T[2].update(r, 1, root[2][c]);
		}
		for (int r: v3[c]) {
			root[3][c] = T[3].update(r, 1, root[3][c]);
		}
	}
}
 
int colour(int ar, int ac, int br, int bc) {
	long long V = (br - ar + 1) * (bc - ac + 1) - (T[0].get(ar, br, root[0][bc]) - T[0].get(ar, br, root[0][ac - 1]));
	long long E = (bc - ac) * (br - ar + 1) - (T[1].get(ar, br, root[1][bc - 1]) - T[1].get(ar, br, root[1][ac - 1])) + (bc - ac + 1) * (br - ar);
 	long long F = 0;
 	if (ar < br) {
 	    E -= (T[2].get(ar, br - 1, root[2][bc]) - T[2].get(ar, br - 1, root[2][ac - 1]));
 	    F = (br - ar) * (bc - ac) - (T[3].get(ar + 1, br, root[3][bc]) - T[3].get(ar + 1, br, root[3][ac]));
 	}
//    if (ar == 2) return 0;
//    if (ar == 3) return 2;
//    if (ar == 5) return 1;
//    if (ar == 1) return 3;
 	return V - E + F;
//    return V - E;
}

• Subtask 1: 0 / 11
• Subtask 2: 12 / 12
• Subtask 3: 0 / 24
• Subtask 4: 0 / 27
• Subtask 5: 0 / 26