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cses grid paths

Nasty Newt answered on March 11, 2022 Popularity 8/10 Helpfulness 2/10

answer cses grid paths

cses grid paths

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#include <iostream>
using namespace std;

const int DIR_LEN = 4;
int dr[DIR_LEN] = {-1, 0, 1, 0};
int dc[DIR_LEN] = {0, 1, 0, -1};
const int PATH_LEN = 48;  // length of all possible paths
int p[PATH_LEN];
const int GRID_SIZE = 9;  
// added border to all four sides so a 7x7 becomes a 9x9
bool onPath[GRID_SIZE][GRID_SIZE];

int tryPath(int pathIdx, int curR, int curC) {
	// Optimization 3
	if ((onPath[curR][curC - 1] && onPath[curR][curC + 1])
		&& (!onPath[curR - 1][curC] && !onPath[curR + 1][curC])) return 0;
	if ((onPath[curR - 1][curC] && onPath[curR + 1][curC])
		&& (!onPath[curR][curC - 1] && !onPath[curR][curC + 1])) return 0;

	if (curR == 7 && curC == 1) {  // reached endpoint before visiting all
		if (pathIdx == PATH_LEN) return 1;
		return 0;
	}

	if (pathIdx == PATH_LEN) return 0;

	int ret = 0;
	onPath[curR][curC] = true;

	// turn already determined:
	if (p[pathIdx] < 4) {
		int nxtR = curR + dr[p[pathIdx]];
		int nxtC = curC + dc[p[pathIdx]];
		if (!onPath[nxtR][nxtC]) ret += tryPath(pathIdx + 1, nxtR, nxtC);
	} 
	// see Java solution for optimization 4 implementation
	else {  // iterate through all four possible turns
		for (int i = 0; i < DIR_LEN; i++) {
			int nxtR = curR + dr[i];
			int nxtC = curC + dc[i];
			if (onPath[nxtR][nxtC]) continue;
			ret += tryPath(pathIdx + 1, nxtR, nxtC);
		}
	}
	// reset and return
	onPath[curR][curC] = false;
	return ret;
}

int main() {
	string line;
	getline(cin, line);

	// convert path to ints
	for (int i = 0; i < PATH_LEN; i++) {
		char cur = line[i];

		if (cur == 'U') p[i] = 0;
		else if (cur == 'R') p[i] = 1;
		else if (cur == 'D') p[i] = 2;
		else if (cur == 'L') p[i] = 3;
		else p[i] = 4; //cur == '?'
	}

	// set borders of grid
	for (int i = 0; i < GRID_SIZE; i++) {
		onPath[0][i] = true;
		onPath[8][i] = true;
		onPath[i][0] = true;
		onPath[i][8] = true;
	}
	// initialize the inside of the grid to be completely empty
	for (int i = 1; i <= 7; i++) {
		for (int j = 1; j <= 7; j++) {
			onPath[i][j] = false;
		}
	}

	int startIdx = 0;
	int startR = 1;
	int startC = 1;  // always start path at (1, 1)
	int ans = tryPath(startIdx, startR, startC);
	cout << ans << endl;
}

Popularity 8/10 Helpfulness 2/10 Language whatever

Source: usaco.guide

Tags: grid whatever

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Contributed on Mar 11 2022

Nasty Newt

0 Answers Avg Quality 2/10

cses grid paths

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