【BZOJ1123】[POI2008]BLO【割顶】

【题目链接】

题意可见discuss。

用Tarjan求割顶，然后对割顶的所有子树求点对个数（前缀和扫一遍），最后把自身和其他点的答案加上。

没注意边，数组开小了。

/* Pigonometry */
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 100005, maxm = 500005;

int n, m, head[maxn], cnt, size[maxn], dfn[maxn], low[maxn], clo;
LL ans[maxn];

struct _edge {
	int v, next;
} g[maxm << 1];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline void add(int u, int v) {
	g[cnt] = (_edge){v, head[u]};
	head[u] = cnt++;
}

inline void tarjan(int x) {
	int sum = 0;
	dfn[x] = low[x] = ++clo;
	size[x] = 1;
	for(int i = head[x]; ~i; i = g[i].next) {
		int v = g[i].v;
		if(!dfn[v]) {
			tarjan(v);
			low[x] = min(low[x], low[v]);
			size[x] += size[v];
			if(low[v] >= dfn[x]) {
				ans[x] += (LL)sum * size[v];
				sum += size[v];
			}
		}
		else low[x] = min(low[x], dfn[v]);
	}
	ans[x] += (LL)sum * (n - sum - 1);
	ans[x] <<= 1;
}

int main() {
	n = iread(); m = iread();
	for(int i = 1; i <= n; i++) head[i] = -1; cnt = 0;

	for(int i = 1; i <= m; i++) {
		int u = iread(), v = iread();
		add(u, v); add(v, u);
	}
	
	tarjan(1);

	for(int i = 1; i <= n; i++) printf("%lld\n", ans[i] + (n - 1) * 2);
	return 0;
}