Submission #395015

#	Time	Username	Problem	Language	Result	Execution time	Memory
395015		soroush	Portals (BOI14_portals)	C++17	100 / 100	915 ms	153136 KiB

portals

//曇り空 のぞいた予感
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int , int> pii;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int maxn = 1010;
const ll mod = 1e9+7;

#define pb push_back
#define endl '\n'
#define dokme(x) cout << x , exit(0)
#define ms(x , y) memset(x , y , sizeof x)
ll pw(ll a, ll b, ll md = mod){ll res = 1;while(b){if(b&1){res=(a*res)%md;}a=(a*a)%md;b>>=1;}return(res);}

int n , m;
char c[maxn][maxn];
int dist[maxn][maxn];
int nie[maxn][maxn];
struct tr{
	int x , y , w;
	friend bool operator < (tr a , tr b){
		return a.w > b.w;
	}
};
pii s , t;
vector < tr > adj[maxn][maxn];
priority_queue < tr > pq;
queue < pii > q;
bool mark[maxn][maxn];

void dijk(){
	ms(dist , 63);
	dist[s.first][s.second] = 0;
	pq.push({s.first, s.second , 0});
	while(pq.size()){
		auto v = pq.top();
		pq.pop();
		if(mark[v.x][v.y])continue;
		if(v.x == t.first and v.y == t.second)dokme(v.w);
		mark[v.x][v.y] = 1;
		for(tr u : adj[v.x][v.y])
			if(dist[u.x][u.y] > v.w + u.w)
				dist[u.x][u.y] = v.w + u.w , pq.push({u.x , u.y , v.w + u.w});
	}
}

pii tmp[maxn][maxn];

int32_t main(){
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin >> n >> m;
	for(int i = 1 ; i <= n ; i ++){
		cin >> (c[i]+1);
		for(int j = 1 ; j <= m ; j ++)
			if(c[i][j] == 'C')
				t = {i , j};
			else if(c[i][j] == 'S')
				s = {i , j};
	}
	ms(nie , -1);
	for(int i = 1 ; i <= n ; i ++)
		for(int j = 1 ; j <= m ; j ++)
			if(c[i][j] == '#')
				nie[i][j] = 0 , q.push({i , j});
	for(int i = 1 ; i <= m ; i ++)
		nie[0][i] = 0 , q.push({0 , i}),
		nie[n + 1][i] = 0 , q.push({n + 1 , i});
	for(int i = 1 ; i <= n ; i ++)
		nie[i][0] = 0 , q.push({i , 0}),
		nie[i][m+1] = 0 , q.push({i , m+1});
	while(q.size()){
		pii v = q.front();
		q.pop();
		if(v.first > 1){
			if(nie[v.first - 1][v.second] == -1)
				nie[v.first - 1][v.second] = nie[v.first][v.second] + 1 , 
				q.push({v.first - 1 , v.second});
		}
		if(v.first < n){
			if(nie[v.first + 1][v.second] == -1)
				nie[v.first + 1][v.second] = nie[v.first][v.second] + 1,
				q.push({v.first + 1 , v.second});
		}
		if(v.second > 1){
			if(nie[v.first][v.second - 1] == -1)
				nie[v.first][v.second - 1] = nie[v.first][v.second] + 1,
				q.push({v.first , v.second-1});
		}
		if(v.second < m){
			if(nie[v.first][v.second + 1] == -1)
				nie[v.first][v.second + 1] = nie[v.first][v.second] + 1,
				q.push({v.first , v.second+1});
		}
	}
	for(int i = 1 ; i <= n ; i ++)
		for(int j = 1 ; j <= m ; j ++){
			if(c[i][j] == '#')continue;
			if(i > 1){
				if(c[i-1][j] != '#')adj[i][j].pb({i-1 , j , 1});
			}
			if(i < n){
				if(c[i+1][j] != '#')adj[i][j].pb({i+1 , j , 1});
			}
			if(j > 1){
				if(c[i][j-1] != '#')adj[i][j].pb({i , j-1 , 1});
			}
			if(j < m){
				if(c[i][j+1] != '#')adj[i][j].pb({i , j+1 , 1});
			}
		}
	
	for(int i = 1 ; i <= n; i ++){
		c[i][0] = '#';
		for(int j = 1 ; j <= m ; j ++){
			if(c[i][j] == '#')continue;
			if(c[i][j-1] == '#')tmp[i][j] = {i , j};
			else tmp[i][j] = tmp[i][j-1];
			adj[i][j].pb({tmp[i][j].first , tmp[i][j].second , nie[i][j]});
		}
	}
	for(int i = 1 ; i <= n; i ++){
		c[i][m+1] = '#';
		for(int j = m ; j >= 1; j --){
			if(c[i][j] == '#')continue;
			if(c[i][j+1] == '#')tmp[i][j] = {i , j};
			else tmp[i][j] = tmp[i][j+1];
			adj[i][j].pb({tmp[i][j].first , tmp[i][j].second , nie[i][j]});
		}
	}
	for(int i = 1 ; i <= m; i ++){
		c[0][i] = '#';
		for(int j = 1 ; j <= n ; j ++){
			if(c[j][i] == '#')continue;
			if(c[j-1][i] == '#')tmp[j][i] = {j , i};
			else tmp[j][i] = tmp[j-1][i];
			adj[j][i].pb({tmp[j][i].first , tmp[j][i].second , nie[j][i]});
		}
	}
	for(int i = 1 ; i <= m; i ++){
		c[n+1][i] = '#';
		for(int j = n ; j >= 1 ; j --){
			if(c[j][i] == '#')continue;
			if(c[j+1][i] == '#')tmp[j][i] = {j , i};
			else tmp[j][i] = tmp[j+1][i];
			adj[j][i].pb({tmp[j][i].first , tmp[j][i].second , nie[j][i]});
		}
	}
	dijk();
	return(0);
}

• Subtask 1: 11 / 11
• Subtask 2: 20 / 20
• Subtask 3: 20 / 20
• Subtask 4: 19 / 19
• Subtask 5: 30 / 30