BZOJ1977 [BJOI2010]次小生成树

题目蓝链

Solution

考虑首先建出一棵最小生成树，然后再枚举所有其他的边。然后直接查询这条边对应在树上的两点之间的链上最大值和次大值，因为要保证严格次小。然后用查询的值更新一下答案

维护一条链上的最大值和次大值，直接倍增就行了

Code

#include <bits/stdc++.h>

using namespace std;

#define squ(x) ((LL)(x) * (x))
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef pair<int, int> pii;

inline int read() {
	int sum = 0, fg = 1; char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') fg = -1;
	for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);
	return fg * sum;
}

const int maxn = 1e5 + 10;
const int maxm = 3e5 + 10;
const LL inf = 0x3f3f3f3f3f3f3f3f;

int n, m;
bool b[maxm];
struct edge {
	int x, y, z;
	bool operator < (const edge &t) const { return z < t.z; }
}e[maxm];

vector<pii> g[maxn];

namespace MST {

	int fa[maxn];
	int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }

	LL kruskal() {
		LL res = 0;
		sort(e + 1, e + m + 1);
		for (int i = 1; i <= n; i++) fa[i] = i;
		for (int i = 1; i <= m; i++) {
			int x = find(e[i].x), y = find(e[i].y);
			if (x == y) continue;
			g[e[i].x].push_back((pii){e[i].y, e[i].z});
			g[e[i].y].push_back((pii){e[i].x, e[i].z});
			fa[x] = y; res += e[i].z; b[i] = 1;
		}
		return res;
	}

}

struct node {
	int x, y;
	node(int _x = -1, int _y = -1): x(_x), y(_y) { }
}Max[maxn][17];

bool cmp(int x, int y) { return x > y; }
int t[5];
node merge(const node &a, const node &b) {
	t[0] = a.x, t[1] = a.y;
	t[4] = b.x, t[3] = b.y;
	sort(t, t + 5, cmp);
	unique(t, t + 5);
	return (node){t[0], t[1]};
}

int fa[maxn][17], d[maxn];

void dfs(int now, int f) {
	fa[now][0] = f, d[now] = d[f] + 1;
	for (int i = 1; i <= 16; i++) {
		fa[now][i] = fa[fa[now][i - 1]][i - 1];
		Max[now][i] = merge(Max[now][i - 1], Max[fa[now][i - 1]][i - 1]);
	}
	for (int i = 0; i < g[now].size(); i++) {
		int son = g[now][i].first, ds = g[now][i].second;
		if (son == f) continue;
		Max[son][0] = (node){ds, -1};
		dfs(son, now);
	}
}

node find(int x, int y) {
	if (d[x] < d[y]) swap(x, y);
	node res = (node){-1, -1};
	for (int i = 16; ~i; i--)
		if (d[fa[x][i]] >= d[y]) res = merge(res, Max[x][i]), x = fa[x][i];
	if (x == y) return res;
	for (int i = 16; ~i; i--)
		if (fa[x][i] != fa[y][i]) {
			res = merge(res, Max[x][i]);
			res = merge(res, Max[y][i]);
			x = fa[x][i], y = fa[y][i];
		}
	res = merge(res, Max[x][0]);
	res = merge(res, Max[y][0]);
	return res;
}

int main() {
#ifdef xunzhen
	freopen("tree.in", "r", stdin);
	freopen("tree.out", "w", stdout);
#endif

	n = read(); m = read();
	for (int i = 1; i <= m; i++) {
		int x = read(), y = read(), z = read();
		e[i] = (edge){x, y, z};
	}

	LL sum = MST :: kruskal(), ans = inf;

	t[4] = -1, dfs(1, 0);

	for (int i = 1; i <= m; i++)
		if (!b[i]) {
			node res = find(e[i].x, e[i].y);
			if (e[i].z > res.x) ans = min(ans, sum + e[i].z - res.x);
			else if (e[i].z > res.y) ans = min(ans, sum + e[i].z - res.y);
		}

	printf("%lld
", ans);

	return 0;
}

Summary

这道题的思路不是很难，结果在码码的过程中出现了很多SB错误，调了我一个小时才调出来...