【二分图染色】ARC 165 C

C - Social Distance on Graph

题意：

思路：

首先考虑一条链的情况，注意到如果两条相邻的边加起来 < x，一定不行

这个结论推广到图也是一样的

同时注意到 x 具有单调性，考虑对 x 二分

在check时进行二分图染色

不合法的情况是相邻的点相同颜色但是边权 < x

Code：

#include 

#define int long long

constexpr int N = 2e5 + 10;
constexpr int M = 1e6 + 10;
constexpr int Inf = 1e9;

std::vector > adj[N]; 

int n, m;
int mi1[N], mi2[N];
int col[N];

bool dfs(int u, int c, int x) {
    col[u] = c;
    for (auto [v, w] : adj[u]) {
        if (col[v] == -1) {
            if (w < x) {
                if (!dfs(v, c ^ 1, x)) return false;
            }
        }else if (col[u] == col[v] && w < x) return false;
    }
    return true;
}
bool check(int mid) {
    memset(col, -1, sizeof(col));
    bool ok = true;
    for (int i = 1; i <= n; i ++) {
        if (col[i] == -1) {
            if (!dfs(i, 0, mid)) {
                ok = false;
                break;
            }
        }
    }
    return ok;
}
void solve() {
    std::cin >> n >> m;
    for (int i = 1; i <= n; i ++) {
        mi1[i] = mi2[i] = 1e18;
    }
    for (int i = 1; i <= m; i ++) {
        int u, v, w;
        std::cin >> u >> v >> w;
        adj[u].push_back({v, w});
        if (mi1[u] > w) {
            mi2[u] = mi1[u];
            mi1[u] = w;
        }else if (mi2[u] > w) {
            mi2[u] = w;
        }
        adj[v].push_back({u, w});
        if (mi1[v] > w) {
            mi2[v] = mi1[v];
            mi1[v] = w;
        }else if (mi2[v] > w) {
            mi2[v] = w;
        }
    }
    int r = 1e18;
    for (int i = 1; i <= n; i ++) {
        r = std::min(r, mi1[i] + mi2[i]);
    }
    int l = 0;
    int ans = 0;
    while (l <= r) {
        int mid = l + r >> 1;
        if (check(mid)) {
            ans = mid;
            l = mid + 1;
        }else {
            r = mid - 1;
        }
    }

    std::cout << ans << "\n";
}
signed main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);

    int t = 1;
    while(t --) {
        solve();
    }
    return 0;
}