2021牛客暑期多校训练营6 J.Defend Your Country 点双连通分量

Posted 2021-08-31 kaka0010

原题链接：https://ac.nowcoder.com/acm/contest/11257/J

目录

• 题意
• 分析
• Code

题意

给定n个点，m条边的无向图，每个点都有$a_i$的权值，定义重要程度为$-1^{联通块大小}\sum ai$，你可以删除任意条边(可以为0)，问最大重要程度是多少。

分析

分两种情况讨论一下

1. 因为题目给定的图是连通图，那么如果n为偶数，直接输出答案。
2. n为奇数的情况下，我们考虑删点，将联通块个数变为偶数。当然为了最优我们只删一个点肯定可以满足。那么讨论这个点怎么删。如果是割点，那么删去这个点之后会有很多个联通块，我们判断一下删完后剩余的联通块是不是都是偶数，如果有一个是奇数，那么不能删；如果不是割点，那么我们都可以删，因为不改变联通性质。最后对两种情况取min就好了。

Code

#include <bits/stdc++.h>
#define lowbit(i) i & -i
#define Debug(x) cout << x << endl
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PII;
const ll INF = 1e18;
const int N = 1e6 + 10;
const int M = 1e6 + 10;
const int MOD = 1e9 + 7;
//-----------------------------------------------------------------IO Template
namespace StandardIO {
    template<typename T>
    inline void read(T &x) {
        x = 0; T f = 1;
        char c = getchar();
        for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = -1;
        for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
        x *= f;
    }

    template<typename T>
    inline void write(T x) {
        if (x < 0) putchar('-'), x *= -1;
        if (x >= 10) write(x / 10);
        putchar(x % 10 + '0');
    }
}
using namespace StandardIO;
//-----------------------------------------------------------------IO Template
int dfn[N], low[N], idx, cut[N], stk[N], tp, ok[N], a[N], siz[N];
vector<int> g[N];
int n, m;
void tarjan(int u, int root) {
    dfn[u] = low[u] = ++idx;
    stk[++tp] = u;
    siz[u] = 1;
    int child = 0;
    for (auto v : g[u]) {
        if (!dfn[v]) {
            tarjan(v, root);
            siz[u] += siz[v];
            low[u] = min(low[u], low[v]);
            if (low[v] >= dfn[u]) {
                child++;
                if (child > 1 || u != root) cut[u] = 1;
                if (siz[v] & 1) ok[u] = 0;
            }
        }
        else
            low[u] = min(low[u], dfn[v]);
    }
}
inline void solve() {
    int T; cin >> T; while (T--) {
        cin >> n >> m;
        ll ans = 0;
        for (int i = 1; i <= n; i++) {
            g[i].clear();
            cut[i] = dfn[i] = low[i] = idx = tp = 0;
            ok[i] = 1;
            cin >> a[i];
            ans += a[i];
        }
        for (int i = 1; i <= m; i++) {
            int u, v; cin >> u >> v;
            g[u].push_back(v);
            g[v].push_back(u);
        }
        if (n % 2 == 0) {
            cout << ans << endl;
            continue;
        }
        for (int i = 1; i <= n; i++) {
            if (!dfn[i]) tarjan(i, i);
        }
        int mn = 1e9;
        for (int i = 1; i <= n; i++) {
            if (cut[i] && ok[i]) mn = min(mn, a[i]);
            if (!cut[i]) mn = min(mn, a[i]);
        }
        cout << ans - 2 * mn << endl;
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
    signed test_index_for_debug = 1;
    char acm_local_for_debug = 0;
    do {
        if (acm_local_for_debug == '$') exit(0);
        if (test_index_for_debug > 20)
            throw runtime_error("Check the stdin!!!");
        auto start_clock_for_debug = clock();
        solve();
        auto end_clock_for_debug = clock();
        cout << "Test " << test_index_for_debug << " successful" << endl;
        cerr << "Test " << test_index_for_debug++ << " Run Time: "
             << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
        cout << "--------------------------------------------------" << endl;
    } while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#else
    solve();
#endif
    return 0;
}