제출 #234001

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
234001		AQT	로봇 (APIO13_robots)	C++14	60 / 100	555 ms	163844 KiB

robots

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
// Problem : APIO '13 P1 - Robots
// Contest : DMOJ
// URL : https://dmoj.ca/problem/apio13p1
// Memory Limit : 128 MB
// Time Limit : 750 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)
 
#include <bits/stdc++.h>
 
using namespace std;
 
int K, N, M;
int dist[10][10][505][505];
vector<pair<int, int>> graph[505][505];
char mp[505][505];
pair<int, int> moves[4] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
pair<int, int> endpos[4][505][505];
vector<pair<int, int>> vec[505*505*10];
int oo = INT_MAX>>2;
 
pair<int, int> fnd(int d, int x, int y){
	if(endpos[d][x][y].first){
		return endpos[d][x][y];
	}
	int r = 0;
	if(mp[x][y] == 'A'){
		r = 3;
	}
	if(mp[x][y] == 'C'){
		r = 1;
	}
	if(mp[x+moves[(d+r)%4].first][y+moves[(d+r)%4].second] == 'x'){
		return endpos[d][x][y] = {x, y};
	}
	return endpos[d][x][y] = fnd((d+r)%4, x+moves[(d+r)%4].first, y+moves[(d+r)%4].second);
}
 
int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin >> K >> M >> N;
	oo = K*M*N;
	for(int l = 1; l<=K; l++){
		for(int r = l; r<=K; r++){
			for(int i = 1; i<=N; i++){
				for(int j = 1; j<=M; j++){
					dist[l][r][i][j] = oo;
				}
			}		
		}
	}
	for(int i = 1; i<=N; i++){
		for(int j = 1; j<=M; j++){
			cin >> mp[i][j];
			if(mp[i][j] > '0' && mp[i][j] <= '9'){
				dist[mp[i][j]-'0'][mp[i][j]-'0'][i][j] = 0;
			}
		}
	}
	for(int i = 0; i<=N+1; i++){
		mp[i][0] = mp[i][M+1] = 'x';
	}
	for(int j = 0; j<=M+1; j++){
		mp[0][j] = mp[N+1][j] = 'x';
	}
	for(int i = 1; i<=N; i++){
		for(int j = 1; j<=M; j++){
			for(int k = 0; k<4; k++){
				graph[i][j].push_back(fnd(k, i, j));
			}
		}
	}
	for(int len = 1; len<=K; len++){
		for(int l = 1; l+len-1<=K; l++){
			int r = l+len-1;
			for(int i = 1; i<=N; i++){
				for(int j = 1; j<=M; j++){
					for(int k = l; k<r; k++){
						dist[l][r][i][j] = min(dist[l][k][i][j] + dist[k+1][r][i][j], dist[l][r][i][j]);
					}
					if(dist[l][r][i][j] != oo){
						//pq.push({dist[l][r][i][j], {i, j}});
						vec[dist[l][r][i][j]].emplace_back(i, j);
					}
				}
			}
			/*
			while(pq.size()){
				auto p = pq.top();
				pq.pop();
				if(dist[l][r][p.second.first][p.second.second] != p.first){
					continue;
				}
				for(auto e : graph[p.second.first][p.second.second]){
					if(dist[l][r][e.first][e.second] > dist[l][r][p.second.first][p.second.second] + 1){
						dist[l][r][e.first][e.second] = dist[l][r][p.second.first][p.second.second] + 1;
						pq.push({dist[l][r][e.first][e.second], {e.first, e.second}});
					}
				}
			}
			*/
			for(int d = 0; d<=oo; d++){
				for(auto p : vec[d]){
					if(dist[l][r][p.first][p.second] != d){
						continue;
					}
					for(auto e : graph[p.first][p.second]){
						if(dist[l][r][e.first][e.second] > d+1){
							dist[l][r][e.first][e.second] = d+1;
							vec[d+1].emplace_back(e.first, e.second);
						}
					}
				}
			}
		}
	}
	int ans = oo;
	for(int i = 1; i<=N; i++){
		for(int j = 1; j<=M; j++){
			ans = min(dist[1][K][i][j], ans);
		}
	}
	/*
	for(int l = 1; l<=K; l++){
		for(int r = l; r<=K; r++){
			cout << l << " " << r << "\n";
			for(int i = 1; i<=N; i++){
				for(int j = 1; j<=M; j++){
					if(dist[l][r][i][j] == INT_MAX>>2){
						cout << 'X';
					}
					else{
						cout << dist[l][r][i][j];
					}
				}
				cout << "\n";
			}
			cout << "\n";
		}
	}
	*/
	cout << (ans == oo ? -1 : ans);
}

컴파일 시 표준 에러 (stderr) 메시지

robots.cpp: In function 'int main()':
robots.cpp:82:60: warning: array subscript is above array bounds [-Warray-bounds]
       dist[l][r][i][j] = min(dist[l][k][i][j] + dist[k+1][r][i][j], dist[l][r][i][j]);
                                                 ~~~~~~~~~~~^
robots.cpp:82:57: warning: array subscript is above array bounds [-Warray-bounds]
       dist[l][r][i][j] = min(dist[l][k][i][j] + dist[k+1][r][i][j], dist[l][r][i][j]);
                                                 ~~~~~~~~^
robots.cpp:82:16: warning: array subscript is above array bounds [-Warray-bounds]
       dist[l][r][i][j] = min(dist[l][k][i][j] + dist[k+1][r][i][j], dist[l][r][i][j]);
       ~~~~~~~~~^

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 30 / 30
• Subtask 4: 0 / 40