USACO Snail Trails 解题报告

这道题就是一道DFS搜索题，我是按照自己习惯的方式进行的，进去之后就mark，出来前unmark。主要是需要看清题，即如果能走的话一直往前走，否则，需要看是否遇到障碍了，如果遇到的是之前走过的点就只能停了。前者的情况可以左转或右转。所以这道题我也保留了方向。

/*
ID: thestor1
LANG: C++
TASK: snail
*/
#include <iostream>
#include <fstream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <climits>
#include <cassert>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>

using namespace std;

bool isIn(int r, int c, int N)
{
return 0 <= r && r < N && 0 <= c && c < N;
}

void search(vector<std::vector<int> > &grid, int r, int c, int d, int cnt, int &largest)
{
// cout<<"[debug]r: "<<r<<", c: "<<c<<endl;
if (cnt > largest)
{
largest = cnt;
// cout<<"[debug]largest: "<<largest<<endl;
}

grid[r][c] = 1;

int N = grid.size();
int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
int nr = r + dx[d], nc = c + dy[d];
if (isIn(nr, nc, N) && !grid[nr][nc])
{
search(grid, nr, nc, d, cnt + 1, largest);
}
else if (!isIn(nr, nc, N) || grid[nr][nc] == 2)
{
int turn[] = {-1, 1};
for (int t = 0; t < 2; ++t)
{
int nd = (d + 4 + turn[t]) % 4;
nr = r + dx[nd];
nc = c + dy[nd];
if (isIn(nr, nc, N) && !grid[nr][nc])
{
search(grid, nr, nc, nd, cnt + 1, largest);
}
}
}

grid[r][c] = 0;
}

int main()
{
ifstream fin("snail.in");
ofstream fout("snail.out");

int N, B;
fin>>N>>B;
std::vector<std::vector<int> > grid(N, std::vector<int>(N, 0));
for (int i = 0; i < B; ++i)
{
char c;
int r;
fin>>c>>r;
// cout<<"r: "<<r<<", c: "<<c<<endl;
grid[r - 1][c - 'A'] = 2;
}
grid[0][0] = 1;

// for (int r = 0; r < N; ++r)
// {
// 	for (int c = 0; c < N; ++c)
// 	{
// 		if (grid[r][c])
// 		{
// 			cout<<'#';
// 		}
// 		else
// 		{
// 			cout<<'.';
// 		}
// 	}
// 	cout<<endl;
// }

int largest = 0;
search(grid, 0, 0, 1, 1, largest);

// cout<<"[debug]after first search: "<<endl;
// for (int r = 0; r < N; ++r)
// {
// 	for (int c = 0; c < N; ++c)
// 	{
// 		if (grid[r][c])
// 		{
// 			cout<<'#';
// 		}
// 		else
// 		{
// 			cout<<'.';
// 		}
// 	}
// 	cout<<endl;
// }

search(grid, 0, 0, 2, 1, largest);
fout<<largest<<endl;

fin.close();
fout.close();
return 0;
}