제출 #1063741

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1063741		danielzhu	Tracks in the Snow (BOI13_tracks)	C++17	93.44 / 100	2021 ms	1048576 KiB

tracks

#include <bits/stdc++.h>
using namespace std;
#define MX 4001
typedef long long LL;
typedef pair<int, int> PR;
typedef tuple<int, int, int> TP;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9 + 5;
const LL MAXV = (LL)1e18; 
using vi = vector<int>;
using vii = vector<vi>;

int H, W;
string md[MX]; //snow meadow
/*
 use Disjoint Sets to solve the problem.
 (1) Use BFS to find all CCs (each with the same footprint letter R or F) and assign each CC with
     an integer ID starting from 0.
 (2) Perform a scan on the meadow to discover connectitvity between adjcent CCs with different letters
 (3) A bipartite graph or tree, then BFS to find the height of the tree which is the answer.
 
 */

vector<unordered_set<int>> adj;


int solve() {
	cin >> H >> W;
	//read the snow meadow info
	for (int i = 0; i < H; i++)
		cin >> md[i]; 
	if (md[0][0] == '.')
		return 0;
  vector<vector<int>> id(H, vector<int>(W, -1));
	int t = 0;
	for (int i = 0; i < H; i++) 
		for (int j = 0; j < W; j++) {
			if (md[i][j] == '.' || id[i][j] != -1)
				continue;
			queue<PR> Q;
			Q.push({i, j});
			id[i][j] = t;
			//BFS to find the CC
      while (!Q.empty()) {
        int r = Q.front().first, c = Q.front().second;
        Q.pop();
				int dr = 0, dc = 1;
        for (int k = 0; k < 4; k++) {
					int nr = r + dr, nc = c + dc;
					if (nr >= 0 && nr < H && nc >= 0 && nc < W && 
					    md[nr][nc] == md[i][j] && id[nr][nc] == -1) {
					  id[nr][nc] = t;
            Q.push({nr, nc});						
					}
					swap(dr, dc);
					dc = -dc;
				}				
			}
      t++;//next CC			
		}

	adj.resize(t);
	
	//scan adjacency between CCs, to build the bipartite graph
	for (int i = 0; i < H; i++)
		for (int j = 0; j < W; j++) {
			if (id[i][j] == -1)
				continue;
			int dr = 0, dc = 1;
			for (int k = 0; k < 4; k++) {
				int r = i + dr, c = j + dc;
				if (r >= 0 && r < H && c >= 0 && c < W && id[r][c] != -1 && id[r][c] != id[i][j] && md[r][c] != md[i][j]) {
					adj[id[i][j]].insert(id[r][c]);
					adj[id[r][c]].insert(id[i][j]);
				}
				swap(dr, dc);
				dc = -dc;
			}
		}
	vector<int> d(t, -1);
	queue<int> Q2;
	Q2.push(0); //the root
	d[0] = 1;
	while (!Q2.empty()) {
		int u = Q2.front();
		Q2.pop();
		for (auto v : adj[u])
			if (d[v] == -1) {
				d[v] = d[u]+1;
				Q2.push(v);
			}
	}
	return *max_element(d.begin(), d.end());
}   


int main() {
	ios::sync_with_stdio(false);
  cin.tie(nullptr);
	cin.exceptions(cin.failbit);
  cout << solve() << endl;
  return 0;
}

• Subtask 1: 30 / 30
• Subtask 2: 63.4375 / 70