BZOJ1098 [POI2007]办公楼biu

题目蓝链

Solution

这题显然就是一道BFS的题，但这道题的边数很多，我们怎么保证是\(\mathcal{O}(n)\)级别的呢

我们只需要维护一个并查集(模拟链表)，表示当前还未访问的点。然后我们从当前节点往所有与当前点连了边的未访问到的节点扩展，每到一个点就在链表删掉当前当前的点。这样就能保证时间复杂度了

至于怎么快速判断两点之间有没有边，我们可以直接把边排好序之后加入到链式前向星，然后在判断的时候直接\(tow point\)判断就可以了

时间复杂度\(\mathcal{O}(n + m)\)

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 = 2e6 + 10;

int n, m;
int Begin[maxn], Next[maxm << 1], To[maxm << 1], e = -1;

inline void add(int x, int y) {
	To[++e] = y, Next[e] = Begin[x], Begin[x] = e;
}

namespace DSU {
	int fa[maxn];
	void init() { for (int i = 1; i <= n + 1; i++) fa[i] = i; }
	int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
	void merge(int x, int y) { x = find(x), y = find(y); if (x != y) fa[x] = y; }
}

bool v[maxn];
int cnt, ans[maxn];

void dfs(int now) {
	v[now] = 1; ++cnt;
	DSU::merge(now, now + 1);
	for (int i = DSU::find(1), p = Begin[now]; i <= n; i = DSU::find(i + 1)) {
		bool fg = 1;
		for (; p + 1 && To[p] <= i; p = Next[p]) if (To[p] == i) { fg = 0; break; }
		if (fg) dfs(i);
	}
}

struct edge {
	int x, y;
	bool operator < (const edge &t) const { return x == t.x ? y > t.y : x > t.x; }
}eg[maxm];

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

	memset(Begin, -1, sizeof Begin);
	n = read(), m = read();
	for (int i = 1; i <= m; i++) {
		int x = read(), y = read();
		eg[i] = (edge){x, y};
	}
	sort(eg + 1, eg + m + 1);
	for (int i = 1; i <= m; i++) add(eg[i].x, eg[i].y), add(eg[i].y, eg[i].x);

	DSU::init();
	for (int i = 1; i <= n; i++)
		if (!v[i]) cnt = 0, dfs(i), ans[++ans[0]] = cnt;

	printf("%d\n", ans[0]);
	sort(ans + 1, ans + ans[0] + 1);
	for (int i = 1; i <= ans[0]; i++) printf("%d%c", ans[i], i < ans[0] ? ' ' : '\n');

	return 0;
}

Summary

复习一下补图并查集，换了一种常数较小的写法