Bipartite Segments CodeForces - 901C (区间二分图计数)

大意: 给定无向图, 无偶环, 每次询问求[l,r]区间内, 有多少子区间是二分图.

无偶环等价于奇环仙人掌森林, 可以直接tarjan求出所有环, 然后就可以预处理出每个点为右端点时的答案.

这样的话区间询问等价于区间求和, 特殊处理一下左右边界的环即可.

要注意同一个点可能属于多个环!!

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif


int n, m, cnt;
vector<int> g[N];
int dfn[N], fa[N], L[N], R[N];
void dfs(int x) {
	dfn[x] = ++*dfn;
	for (int y:g[x]) {
		if (dfn[y]) {
			if (dfn[y]<dfn[x]||y==x) continue;
			fa[x] = y, ++cnt;
			L[cnt]=R[cnt]=x;
			for (; y!=x; y=fa[y]) {
				L[cnt] = min(L[cnt], y);
				R[cnt] = max(R[cnt], y);
			}
		}
		else fa[y]=x, dfs(y);
	}
}

int f[N];
ll sum[N];

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,m) {
		int u, v;
		scanf("%d%d", &u, &v);
		g[u].pb(v),g[v].pb(u);
	}
	REP(i,1,n) if (!dfn[i]) dfs(i);
	REP(i,1,cnt) f[R[i]] = max(f[R[i]], L[i]);
	REP(i,1,n) { 
		f[i]=max(f[i],f[i-1]);
		sum[i]=sum[i-1]+(i-f[i]);
	}
	int q;
	scanf("%d", &q);
	REP(i,1,q) {
		int l, r;
		scanf("%d%d", &l, &r);
		int pos = lower_bound(f+1,f+1+n,l)-f;
		if (pos>r) {
			printf("%lld\n", (ll)(r-l+1)*(r-l+2)/2);
		}
		else {
			ll ans = sum[r]-sum[pos-1];
			ll d = pos-l;
			printf("%lld\n", ans+d*(1+d)/2);
		}
	}
}