題目連結

思路

• 比較直接，就是用vector去儲存每個連通區域，連通區域內的畫素點用unordered_set儲存，方便之後的查詢
• 每次新來一個畫素點的時候，就在所有列表中找他是否屬於某個鄰域，如果是的話，則將所有的這些鄰域進行合併。否則就新建一個unordered_set，儲存這個新的land即可。
• 總結：滴滴的題目主要是有了思路就得立馬寫，流程比較複雜，但是一些技巧性的演算法可能涉及的不太多。

程式碼

#include <cstdlib>
#include <string>
#include <iostream>
#include <fstream>
#include <vector>
#include <sstream>
#include <unordered_map>
#include <unordered_set>
#include <map>
#include <set>
#include <stdio.h>
#include <numeric>
#include <algorithm>
#include <functional>
#include <stack>
#include <queue>
#include <cmath>
#include <assert.h>
#include <vector>
#include <memory>
#include <memory.h>
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
typedef unsigned char uchar;

//#define G_DEBUG

int dirs4[5][2] = { -1, 0, 1, 0, 0, 0, 0, 1, 0, -1 };

struct pairhash {
public:
    template <typename T, typename U>
    std::size_t operator()(const std::pair<T, U> &x) const
    {
        return std::hash<T>()(x.first) ^ std::hash<U>()(x.second);
    }
};


int main()
{

#ifdef G_DEBUG
    // 除錯使用
    int N = 2;
    string str1 = "abab";
    string str2 = "ababab";
#else
    int rows = 0, cols = 0;
    int count = 0;

    cin >> rows >> cols >> count;
#endif
    vector<unordered_set<pair<int, int>, pairhash>> pairs;
    int curr_r = 0, curr_c = 0;
    int result = 0;
    for (int i = 0; i < count; i++)
    {
        cin >> curr_r >> curr_c;
        if ((curr_r < 0 || curr_r >= rows)
            || (curr_c < 0 || curr_c >= cols))
        {
            if (i == 0)
                cout << pairs.size();
            else
                cout << " " << pairs.size();
            continue;
        }


        vector<int> need_cmb;
        for (int k = 0; k < pairs.size(); k++)
        {
            bool found = false;
            auto& curr_pair = pairs[k];
            for (int j = 0; j < 5; j++)
            {
                int tmp_r = curr_r + dirs4[j][0];
                int tmp_c = curr_c + dirs4[j][1];
                if ((tmp_r < 0 || tmp_r >= rows)
                    || (tmp_c < 0 || tmp_c >= cols))
                    continue;
                if (curr_pair.find(pair<int, int>(tmp_r, tmp_c)) != curr_pair.end())
                {
                    curr_pair.insert(pair<int, int>(curr_r, curr_c));
                    found = true;
                    break;
                }
            }
            if (found)
            {
                need_cmb.push_back( k );
            }
        }

        if (need_cmb.empty())
        {
            pairs.resize( pairs.size()+1 );
            pairs.back().insert( pair<int,int>(curr_r, curr_c) );
        }
        else if (need_cmb.size() > 1)
        {
            for (int k = need_cmb.size() - 1; k >= 1; k--)
            {
                for (auto c : pairs[need_cmb[k]])
                {
                    pairs[need_cmb[0]].insert( c );
                }
                pairs.erase( pairs.begin()+need_cmb[k] );
            }
        }


        if (i == 0)
            cout << pairs.size();
        else
            cout << " " << pairs.size();
    }
    cout << endl;


    system("pause");
    return 0;
}