Codeforces 901C Bipartite Segments（Tarjan + 二分）

题目链接  Bipartite Segments

题意  给出一个无偶环的图，现在有$q$个询问。求区间$[L, R]$中有多少个子区间$[l, r]$

　　满足$L <= l <= r <= R$，并且一个只包含$l$到$r$这些点的无向图为二分图。

因为整张图没有偶环，所以在这道题中如果某个无向图没有环，那个这个无向图就是二分图

Tarjan求出每个环的标号最小点和标号最大点。

令$f[i]$为能保证$[i, j]$这个区间构成的图都是二分图的$j$的最大值，则$f[i]$是不下降的

当询问区间$[L, R]$的时候，即求$\begin{equation*}\sum_{i=L}^Rmin(f(i), R) - i + 1\end{equation*}$

二分然后分类讨论即可

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)

typedef long long LL;

const int N = 3e5 + 10;

int f[N], dfn[N], stk[N];
int n, m, ti = 0;
int q, top;
LL s[N];
vector <int> v[N];

void dfs(int x, int fa){
	dfn[x] = ++ti;
	stk[++top] = x;
	for (auto u : v[x]){
		if (u == fa) continue;
		if (dfn[u]){
			if (dfn[u] < dfn[x]){
				int mx = u, mi = u;
				for (int p = top; stk[p] != u; --p){
					mi = min(mi, stk[p]);
					mx = max(mx, stk[p]);
				}
				f[mi] = mx;
			}
		}
		else dfs(u, x);
	}
	--top;
}

int main(){

	scanf("%d%d", &n, &m);
	rep(i, 1, m){
		int x, y;
		scanf("%d%d", &x, &y);
		v[x].push_back(y);
		v[y].push_back(x);
	}

	rep(i, 1, n) if (!dfn[i]) dfs(i, 0);
	rep(i, 1, n) if (f[i]) --f[i]; else f[i] = n;
	dec(i, n - 1, 1) f[i] = min(f[i], f[i + 1]);
	rep(i, 1, n) s[i] = s[i - 1] + f[i];

	scanf("%d", &q);
	while (q--){
		int x, y;
		scanf("%d%d", &x, &y);
		if (f[x] > y){
			LL num = y - x + 1;
			printf("%lld\n", 0ll - 1ll * (x + y) * num / 2 + 1ll * y * num + num);
			continue;
		}

		int l = x, r = y;
		while (l + 1 < r){
			int mid = (l + r) >> 1;
			if (f[mid] <= y) l = mid; else r = mid - 1;
		}

		int t;
		if (f[r] <= y) t = r; else t = l;

		int u = t + 1;
		LL num = y - x + 1;
		LL ans = num - 1ll * (x + y) * num / 2;

		LL et = y - u + 1;
		ans = ans + (s[t] - s[x - 1]);
		if (et > 0) ans = ans + 1ll * y * et;
		printf("%lld\n", ans);

	}
	return 0;
}