洛谷 [P3496] BLO

割点

首先 tarjan 求割点,
对于不是割点的点, 答案是 2 * (n-1) 有序，所以要乘 2
对于是割点的点, 答案是删去该点后所有连通块的个数加上 n-1 在乘 2

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
const int MAXN = 1000005;
ll ans[MAXN];
int head[MAXN], n, m, nume, dfn[MAXN], low[MAXN], ind, siz[MAXN];
bool f[MAXN];
struct edge{
	int to, nxt;
}e[MAXN];
void adde(int from, int to) {
	e[++nume].to = to;
	e[nume].nxt = head[from];
	head[from] = nume;
}
int init() {
	int rv = 0, fh = 1;
	char c = getchar();
	while(c < '0' || c > '9') {
		if(c == '-') fh = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		rv = (rv<<1) + (rv<<3) + c - '0';
		c = getchar();
	}
	return fh * rv;
}
void tarjan(int u) {
	dfn[u] = low[u] = ++ind;siz[u] = 1;
	int flag = 0, sum = 0;
	for(int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if(!dfn[v]) {
			tarjan(v);
			siz[u] += siz[v];
			low[u] = min(low[u], low[v]);
			if(low[v] >= dfn[u]) {
				flag++;
				ans[u] += (ll) siz[v] * (n - siz[v]);
				sum += siz[v];
				if(u != 1 || flag > 1) {
					f[u] = 1;
				}
			}
		}else low[u] = min(low[u], dfn[v]);
	}
	if(f[u]) ans[u] += (ll) (n - sum - 1) * (sum + 1) + n - 1;
	else ans[u] = 2 * n - 2;
}
int main() {
	n = init(); m = init();
	for(int i = 1; i <= m; i++) {
		int u = init(), v = init();
		if(u == v) continue;
		adde(u, v); adde(v, u);
	}
	tarjan(1);
	for(int i = 1; i <= n; i++) printf("%lld\n", ans[i]);
	return 0;
}