图论习题

1.树的搜索的板子题。每次修改的时候在子树根节点标记，并且向下传递dfs即可

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define ll long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
int _,n,m;
const int N = 2e5 + 10,inf = 1e9;
const int mod = 1e9 + 7;
int c[N];
vector<int> g[N];

void dfs(int u,int fa){
    c[u] += c[fa];
    for(int v : g[u])
       if(v != fa)
         dfs(v,u);
}

int main()
{
    cin >> n >> m;
    for (int i = 1,u,v; i < n; i ++ ){
        scanf("%d %d", &v, &u);
        g[u].push_back(v);
        g[v].push_back(u);
    }
    
    for (int i = 1,a,b; i <= m; i ++ ){
        scanf("%d %d", &a, &b);
        c[a] += b;
    }
    dfs(1,0);
    for (int i = 1; i <= n; i ++ ) cout << c[i] << " ";
    cout << endl;
    return 0;
}

 2.首先vector构图，然后通过两种不同的颜色计分，如果分数为0则平衡 + 1

然后dfs从第一个根节点开始搜索

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
#define cy cout << "YES" << endl
#define cn cout << "NO" << endl
int _,n,m,cnt;
const int N = 4e3 + 10,inf = 1e9;
const int mod = 1e9 + 7;
int color[N],a[N];
vector<int> g[N];

int dfs(int u)
{
    int sum = color[u];//记录根节点的分数
    for (int i = 0; i < g[u].size(); i ++ ) sum += dfs(g[u][i]); //搜索以u为根节点的每个儿子的分数
    if(sum == 0) cnt ++; //如果平衡，数量 + 1
    return sum;
}

signed main()
{
    cin >> _;
    while(_ -- ){
        for (int i = 1; i <= n; i ++ ) g[i].clear();
        char c;
        cin >> n;
        cnt = 0;
        for (int i = 2; i <= n; i ++ ) cin >> a[i];
    
        //g[1].push_back(1);
        for (int i = 2; i <= n; i ++ ){
            g[a[i]].push_back(i);
        }
        for (int i = 1; i <= n; i ++ ){
            cin >> c;
            if(c == 'W') color[i] = -1;
            else color[i] = 1;
        }
        
        dfs(1);//根节点是1
        cout << cnt << endl;
    }
    return 0;
}