Solution 「CF 1023F」Mobile Phone Network

\(\mathcal{Description}\)

  Link.

  有一个 \(n\) 个结点的图，并给定 \(m_1\) 条无向带权黑边，\(m_2\) 条无向无权白边。你需要为每条白边指定边权，最大化其边权和，并保证 \(m_2\) 条边都在最小生成树中。

  \(n,m_1,m_2\le5\times10^5\)。

\(\mathcal{Solution}\)

  先保证在 \(\text{MST}\) 中的限制——指定所有边权为 \(0\)。并求出此时的 \(\text{MST}\)。显然最优情况下，\(\text{MST}\) 的形态和现在一样。

  那么对于每一条不在 \(\text{MST}\) 上的黑边，相当于限制了一条树上路径的最大值。从小到大枚举这样的边，每个点维护一个指针（整体上就是一个并查集）指向第一个未被限制到父亲边权的祖先，暴力跳指针统计答案即可。

  复杂度 \(\mathcal O\left((m_1+m_2)\log(m_1+m_2)\right)\)。

\(\mathcal{Code}\)

#include <cstdio>
#include <algorithm>

#define ww first
#define uu second.first
#define vv second.second

const int MAXN = 5e5, MAXM = 1e6;
int n, m1, m2, m, fa[MAXN + 5];
int ecnt, head[MAXN + 5], trf[MAXN + 5], trc[MAXN + 5], dep[MAXN + 5];
std::pair<int, std::pair<int, int> > eset[MAXM + 5];
bool inmst[MAXM + 5];

struct Edge { int to, cst, nxt; } graph[MAXN * 2 + 5];

inline int rint () {
	int x = 0; char s = getchar ();
	for ( ; s < '0' || '9' < s; s = getchar () );
	for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
	return x;
}

inline void link ( const int s, const int t, const int c ) {
	graph[++ ecnt] = { t, c, head[s] };
	head[s] = ecnt;
}

inline void init () { for ( int i = 1; i <= n; ++ i ) fa[i] = i; }

inline int find ( const int x ) { return x ^ fa[x] ? fa[x] = find ( fa[x] ) : x; }

inline bool unite ( int x, int y ) {
	return ( x = find ( x ) ) ^ ( y = find ( y ) ) ? fa[x] = y, true : false;
}

inline void DFS ( const int u ) {
	for ( int i = head[u], v; i; i = graph[i].nxt ) {
		if ( ( v = graph[i].to ) ^ trf[u] ) {
			trf[v] = u, trc[v] = graph[i].cst, dep[v] = dep[u] + 1;
			DFS ( v );
		}
	}
}

int main () {
	n = rint (), m = ( m1 = rint () ) + ( m2 = rint () );
	init ();
	for ( int i = 1; i <= m1; ++ i ) {
		eset[i].uu = rint (), eset[i].vv = rint ();
		eset[i].ww = 0;
	}
	for ( int i = m1 + 1; i <= m; ++ i ) {
		eset[i].uu = rint (), eset[i].vv = rint ();
		eset[i].ww = rint ();
	}
	std::sort ( eset + 1, eset + m + 1 );
	for ( int i = 1, cnt = 0; i <= m; ++ i ) {
		if ( unite ( eset[i].uu, eset[i].vv ) ) {
			inmst[i] = true;
			link ( eset[i].uu, eset[i].vv, eset[i].ww );
			link ( eset[i].vv, eset[i].uu, eset[i].ww );
			if ( ++ cnt == n - 1 ) break;
		}
	}
	DFS ( 1 ), init ();
	long long ans = 0; int limited = 0;
	for ( int i = m1 + 1; i <= m; ++ i ) {
		if ( inmst[i] ) continue;
		int u = find ( eset[i].uu ), v = find ( eset[i].vv ), w = eset[i].ww;
		while ( u ^ v ) {
			if ( dep[u] < dep[v] ) u ^= v ^= u ^= v;
			if ( ! trc[u] ) ans += w, ++ limited;
			int t = find ( trf[u] );
			unite ( u, t ), u = t;
		}
	}
	printf ( "%lld\n", limited == m1 ? ans : -1 );
	return 0;
}