寻找俄罗斯方块
寻找俄罗斯方块:
思路:记忆化BFS。直接匹配四种旋转情况,可直接AC
代码:
#include <iostream>
#include <queue>
using namespace std;
typedef pair<int,int> PII;
int n, m;
int dirx[4] = {-1, 0, 1, 0};
int diry[4] = {0, -1, 0 ,1};
int main()
{
cin >> m >> n;
vector<vector<int>> mp(m, vector<int> (n));
vector<vector<bool>> st(m, vector<bool> (n, false));
for(int i = 0; i < m; i++)
for(int j = 0; j < n; j++)
cin >> mp[i][j];
for(int i = 0; i < m; i++)
for(int j = 0; j < n; j++)
{
if(mp[i][j] && !st[i][j])
{
st[i][j] = true;
int up = 0, down = 0, left = 0, right = 0;
queue<PII> q;
q.push({i, j});
while(q.size())
{
auto t = q.front(); q.pop();
int x = t.first, y = t.second;
for(int k = 0; k < 4; k++)
{
int nex = x + dirx[k], ney = y + diry[k];
if(nex >= 0 && nex < m && ney >= 0 && ney < n && !st[nex][ney] && mp[nex][ney])
{
st[nex][ney] = true;
if(k == 0) up++;
if(k == 1) left++;
if(k == 2) down++;
if(k == 3) right++;
q.push({nex, ney});
}
}
}
//做匹配
if(down == 2 && right == 1 && left == 0 && up == 0)
cout << i * n + j << " " << i * n + j + 1 << " " << (i + 1) * n + j << " " << (i + 2) * n + j << endl;
if(down == 1 && right == 2 && left == 0 && up == 0)
{
if(mp[i+1][j])
cout << i * n + j << " " << (i + 1) * n + j << " " << (i + 1) * n + j + 1 << " " << (i + 1) * n + j + 2 << endl;
else
cout << i * n + j << " " << i * n + j + 1 << " " << i * n + j + 2 << " " << (i + 1) * n + j + 2 << endl;
}
if(down == 2 && left == 1 && right == 0 && up == 0)
cout << i * n + j << " " << (i + 1) * n + j << " " << (i + 2) * n + j - 1 << " " << (i + 2) * n + j << endl;
}
}
return 0;
} 