【UOJ #14】【UER #1】DZY Loves Graph

http://uoj.ac/problem/14

题解很好的~

不带路径压缩的并查集能保留树的原本形态。

按秩合并并查集可以不用路径压缩，但是因为此题要删除，如果把深度当为秩的话不好更新秩的值，所以把子树大小当为秩。

合并直接合并，删除直接删除，每条边只会被添加进树一次，至多被删除一次。

离线特殊考虑一下return的情况就可以了QwQ

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 300003;
const int M = 500003;
int in() {
	int k = 0; char c = getchar();
	for(; c < '0' || c > '9'; c = getchar());
	for(; c >= '0' && c <= '9'; c = getchar())
		k = k * 10 + c - 48;
	return k;
}

int fa[N], sz[N], q[M], tail = 0, intree[M], n, m, cnt = 0;

ll ans[M], treesum = 0;

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

void merge(int x, int y, int len) {
	x = find(x); y = find(y);
	q[++tail] = len;

	if (x == y)
		intree[len] = -1;
	else {
		++cnt;
		treesum += (ll) len;
		if (sz[x] > sz[y]) swap(x, y);
		intree[len] = x;
		fa[x] = y;
		sz[y] += sz[x];
		while (fa[y] != y) {
			y = fa[y];
			sz[y] += sz[x];
		}
	}
}

void del() {
	int x = q[tail--];
	if (intree[x] == -1) return;

	--cnt;
	treesum -= x;
	x = intree[x];

	int y = fa[x];
	sz[y] -= sz[x];
	while (fa[y] != y) {
		y = fa[y];
		sz[y] -= sz[x];
	}
	fa[x] = x;
}

char c[20];
struct node {
	char op;
	int x, y;
} Q[M];

int main() {
	n = in(); m = in();
	for(int i = 1; i <= m; ++i) {
		scanf("%s", c);
		if ((Q[i].op = c[0]) == 'R') continue;
		Q[i].x = in();
		if (c[0] == 'A')
			Q[i].y = in();
	}

	for(int i = 1; i <= n; ++i)
		fa[i] = i, sz[i] = 1;

	for(int i = 1; i <= m; ++i) {
		switch (Q[i].op) {
			case 'A':
				merge(Q[i].x, Q[i].y, i);
				printf("%lld\n", ans[tail] = cnt < n - 1 ? 0 : treesum);
			break;
			case 'D':
				if (Q[i + 1].op == 'R') {
					printf("%lld\n", ans[tail - Q[i].x]);
					printf("%lld\n", ans[tail]);
				} else {
					for(int j = 1; j <= Q[i].x; ++j)
						del();
					printf("%lld\n", ans[tail]);
				}
			break;
			case 'R':
				if (Q[i - 1].op == 'A') {
					del();
					printf("%lld\n", ans[tail]);
				}
			break;
		}
	}

	return 0;
}