牛客算法周周练E地、颜色、魔法

题目链接

一个不难的BFS染色题，难点在于开空间。

题目给的范围是n*m<=1e6

如果直接在堆里开空间必须开到1e6*1e6，显然爆了，如果在栈里边开就不能再全局跑BFS，于是就用vector动态开空间。

AC代码：

/*
 * @Author: hesorchen
 * @Date: 2020-07-03 17:05:01
 * @LastEditTime: 2020-08-11 19:34:13
 * @Description: https://hesorchen.github.io/
 */
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define endl '\n'
#define PI acos(-1)
#define PB push_back
#define ll long long
#define INF 0x3f3f3f3f
#define mod 1000000007
#define pll pair
#define lowbit(abcd) (abcd & (-abcd))
#define max(a, b) ((a > b) ? (a) : (b))
#define min(a, b) ((a < b) ? (a) : (b))

#define IOS                      \
    ios::sync_with_stdio(false); \
    cin.tie(0);                  \
    cout.tie(0);
#define FRE                              \
    {                                    \
        freopen("in.txt", "r", stdin);   \
        freopen("out.txt", "w", stdout); \
    }

inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}
//head==============================================================================

vector<char>mp[1000010];
vector<int>vis[1000010];

struct node
{
    ll x, y;
}temp1;
queue<node>q;
int mov[4][2]={ 0, 1, 0, -1, 1, 0, -1, 0 };
ll n, m;
void bfs(ll x, ll y)
{
    q.push(node{ x, y });
    vis[x][y]=1;
    while (q.size())
    {
        temp1=q.front();
        q.pop();
        ll x=temp1.x;
        ll y=temp1.y;
        for (int i=0;i<4;i++)
        {
            ll xx= x+mov[i][0];
            ll yy= y+mov[i][1];
            if (xx>=1&&xx<=n&&yy>=1&&yy<=m&&!vis[xx][yy]&&mp[xx][yy]=='.')
            {
                vis[xx][yy]=1;
                q.push(node{ xx, yy });
            }
        }
    }
}
int main()
{
    cin>>n>>m;
    getchar();
    char temp;
    for (int i=1;i<=n;i++)
    {
        mp[i].push_back('?');
        vis[i].push_back(0);
        for (int j=1;j<=m;j++)
        {
            scanf("%c", &temp);
            mp[i].push_back(temp);
            vis[i].push_back(0);
        }
        getchar();
    }
    for (int i=1;i<=n;i++)
    {
        if (mp[i][1]=='.'&&!vis[i][1])
        {
            bfs(i, 1);
        }
        if (mp[i][m]=='.'&&!vis[i][m])
        {
            bfs(i, m);
        }
    }
    for (int j=1;j<=m;j++)
    {
        if (mp[1][j]=='.'&&!vis[1][j])
        {
            bfs(1, j);
        }
        if (mp[n][j]=='.'&&!vis[n][j])
        {
            bfs(n, j);
        }
    }
    ll ans =0;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
            ans+=(vis[i][j]^1);
    cout<<ans<<endl;
    return 0;
}

/*
5 5
...#.
.##..
.#.#.
.###.
.....
 */