智算之道 - 情报战 (dfs + 记忆化)

题意:

思路:
每知道一个数，我们就能知道与这个数相对应的其他数，在知道其他数的同时我们又知道了，其他数所对应的其他数，所以一直这样递归下去，再加上个记忆化就可以了

AC代码

#include <bits/stdc++.h>
inline long long read(){char c = getchar();long long x = 0,s = 1;
while(c < '0' || c > '9') {if(c == '-') s = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x*10 + c -'0';c = getchar();}
return x*s;}
using namespace std;
#define NewNode (TreeNode *)malloc(sizeof(TreeNode))
#define Mem(a,b) memset(a,b,sizeof(a))
#define lowbit(x) (x)&(-x)
const int N = 3e5 + 5;
const long long INFINF = 0x7f7f7f7f7f7f7f;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-7;
const int mod = 1e9+7;
const double II = acos(-1);
const double PP = (II*1.0)/(180.00);
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> piil;
vector<int> v[N];
int vis[N],sum;
void solve(int num)
{
    for(int i = 0;i < v[num].size();i++)
    {
        if(!vis[v[num][i]])
        {
            vis[v[num][i]] = 1;sum++;
            solve(v[num][i]);
        }
    }
}
signed main()
{
    std::ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    //    freopen("input.txt","r",stdin);
    //    freopen("output.txt","w",stdout);
    int n,m;
    cin >> n >> m;
    while(m--)
    {
        int a;
        cin >> a;
        if(a == 1)
        {
            int num;
            cin >> num;
            if(!vis[num])
            {
                vis[num] = 1,sum++;
                solve(num);
            }
        }
        else
        {
            int num1,num2;
            cin >> num1 >> num2;
            if((vis[num1] && !vis[num2]) || (!vis[num1] && vis[num2]))
            {
                if(vis[num1])
                {
                    vis[num2] = 1;sum++;
                    solve(num2);
                }
                else
                {
                    vis[num1] = 1;sum++;
                    solve(num1);
                }
            }
            else if(!vis[num1] && !vis[num2])
                v[num1].push_back(num2),v[num2].push_back(num1);
        }
        cout << sum << endl;
    }
}