C - Infected Tree

一个简单的dp ，统计每一个结点包括自身和所有下属结点的个数sum，并统计每一个结点感染后可以保存最多多少个结点ss表示，如果他的分支为0就是0为1就是他的儿子结点的sum-1，如果为2则是max(a[pace[0]].sum - 1 + a[pace[1]].ss, a[pace[1]].sum - 1 + a[pace[0]].ss)；然后就是一些小细节的实现了

代码：

#include<bits/stdc++.h>
#define AC return 0;
#define int long long
using namespace std;
typedef pair<int, int> pii;
const int maxx = 300005;
const int mod = 1e9 + 7;
int n, m, t;
struct node {
    int sum;
    vector<int>next;
    bool vis;
    int deep;
    int ss;
    node() {
        deep = 0;
        vis = 0;
        sum = 0;
        ss = 0;
        next.clear();
    }
} a[maxx];
int dfs(int k) {

a[k].vis = 1;
    vector<int>pace;
    int len = a[k].next.size();
    for (int i = 0; i < len; i++)if(!a[a[k].next[i]].vis)pace.push_back(a[k].next[i]);
    if (a[k].next.size() == 1 && k != 1) {
        a[k].sum = 1;
        a[k].deep = 1;
        return a[k].sum;
    } else {
        a[k].sum = 1;
        int p = a[k].next.size();
        for (int i = 0; i < p ; i++) {
            if (!a[a[k].next[i]].vis)a[k].sum += dfs(a[k].next[i]);
        }
    }

if (pace.size() == 0) {
        a[k].ss = 0;
    } else if (pace.size() == 1) {
        a[k].ss = a[pace[0]].sum - 1;
    } else {
        a[k].ss = max(a[pace[0]].sum - 1 + a[pace[1]].ss, a[pace[1]].sum - 1 + a[pace[0]].ss);
    }
    return a[k].sum;
}
int ans = 0;
signed  main() {
    cin >> t;
    while (t--) {
        cin >> n;
        ans = 0;
        for (int i = 0; i <= n; i++) {
            a[i].next.clear();
            a[i].sum = 0;
            a[i].vis = 0;
            a[i].deep = 0;
            a[i].ss = 0;
        }
        for (int i = 1; i < n; i++) {
            int x, y;
            cin >> x >> y;
            a[x].next.push_back(y);
            a[y].next.push_back(x);
        }
        dfs(1);
        cout << a[1].ss << '\n';
    }
    AC
}

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