传送门

题意：

给你一个矩阵，矩阵中的每一个数的二进制串可以表示出该位置周围的围墙信息，让你求出所有的房间大小。

思路：

建图，搜索，一气呵成。

#include<iostream>
#include<string>
#include<map>
#include<bits/stdc++.h>
#include<iomanip>
#include<algorithm>
#define ll long long
using namespace std;
int g[4050][4050];
int bit[10];
void get_bit(int a)
{int cnt = 0;while(a){bit[++cnt] = a%2;a = a>>1; }
}
int h[4] = {2,0,-2,0},l[4] = {0,2,0,-2};
int n,m;
int ans[1000010];
int maxx = 0;
int exchange(int x)
{if(!x)return x;int t = x/abs(x);return t;
}
void dfs(int i, int j)
{for(int k = 0; k < 4; k++){if(!g[i+h[k]][j+l[k]] && !g[i+exchange(h[k])][j+exchange(l[k])] && i+h[k] <= (n-1)*2+4 && i+h[k] >= 4 && j+l[k] <= (m-1)*2+4 && j+l[k] >= 4){maxx++;g[i+h[k]][j+l[k]] = 1;dfs(i+h[k], j+l[k]);}}
}
int main()
{cin>>n>>m;for(int i = 4; i <= (n-1)*2+4; i+=2){for(int j = 4; j <= (m-1)*2+4; j+=2){int a;cin>>a;get_bit(a);for(int k = 4; k >= 1; k--){if(k == 4 && bit[k])g[i-1][j] = 1;else if(k == 3 && bit[k])g[i][j+1] = 1;else if(k == 2 && bit[k])g[i+1][j] = 1;else if(k == 1 && bit[k])g[i][j-1] = 1;}memset(bit,0,sizeof(bit));}}int num = 0;for(int i = 4; i <= (n-1)*2+4; i+=2){for(int j = 4; j <= (m-1)*2+4; j+=2){maxx = 1;if(!g[i][j]){g[i][j] = 1;dfs(i,j);ans[++num] = maxx;}}}sort(ans+1,ans+1+num);for(int i = num; i >= 1; i--)cout<<ans[i]<<" ";
}

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题意：

思路：

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