Codeforces 901C Bipartite Segments

因为图中只存在奇数长度的环，所以它是个只有奇数环的仙人掌，每条边只属于一个环。

那么我们能把所有环给扣出来，所以我们询问的区间不能包含每个环里的最大值和最小值，

这个东西能用dfs直接扣，找最大值和最小值能用倍增，或者直接tarjan扣出来就好。然后

我们可以处理出每个点开始往右延伸的最大位置，求答案能离线线段树（我是这么写的），

但有一点就是这个数组是个单调的数组，所以我们能二分出那个刚好超过R的位置，直接求就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long

using namespace std;

const int N = 3e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

int n, m, q, to[N], f[N][20], mx[N][20], mn[N][20], depth[N];
bool vis[N];
LL ans[N];
vector<int> G[N];
vector<PII> qus[N];

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
LL a[N << 2], lazy[N << 2];
inline void pull(int rt) {
    a[rt] = a[rt << 1] + a[rt << 1 | 1];
}
inline void push(int rt, int l, int mid, int r) {
    if(lazy[rt]) {
        a[rt << 1] += lazy[rt] * (mid - l + 1);
        a[rt << 1 | 1] += lazy[rt] * (r - mid);
        lazy[rt << 1] += lazy[rt];
        lazy[rt << 1 | 1] += lazy[rt];
        lazy[rt] = 0;
    }
}
void update(int L, int R, LL val, int l, int r, int rt) {
    if(l >= L && r <= R) {
        a[rt] += val * (r - l + 1);
        lazy[rt] += val;
        return;
    }
    int mid = l + r >> 1;
    push(rt, l, mid, r);
    if(L <= mid) update(L, R, val, lson);
    if(R > mid)  update(L, R, val, rson);
    pull(rt);
}
LL query(int L, int R, int l, int r, int rt) {
    if(l >= L && r <= R) return a[rt];
    int mid = l + r >> 1;
    push(rt, l, mid, r);
    if(R <= mid) return query(L, R, lson);
    else if(L > mid) return query(L, R, rson);
    else return query(L, R, lson) + query(L, R, rson);
}


void dfs(int u, int fa) {
    vis[u] = true;
    depth[u] = depth[fa] + 1;
    f[u][0] = fa;
    mx[u][0] = u, mn[u][0] = u;
    for(int i = 1; i < 20; i++) {
        f[u][i] = f[f[u][i - 1]][i - 1];
        mx[u][i] = max(mx[u][i - 1], mx[f[u][i - 1]][i - 1]);
        mn[u][i] = min(mn[u][i - 1], mn[f[u][i - 1]][i - 1]);
    }
    for(auto& v : G[u]) {
        if(v == fa || vis[v]) continue;
        dfs(v, u);
    }
}

void getVal(int u, int v) {
    int MX = v, MN = v;
    for(int i = 19; i >= 0; i--) {
        if(depth[u] - depth[v] >= (1 << i)) {
            MX = max(MX, mx[u][i]);
            MN = min(MN, mn[u][i]);
            u = f[u][i];
        }
    }
    to[MN] = MX - 1;
}

void getTo(int u, int fa) {
    vis[u] = true;
    for(auto& v : G[u]) {
        if(v == fa) continue;
        else if(!vis[v]) getTo(v, u);
        else if(vis[v]) {
            if(depth[u] > depth[v]) getVal(u, v);
        }
    }
}

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) to[i] = n;
    for(int i = 1; i <= m; i++) {
        int u, v; scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    for(int i = 1; i <= n; i++)
        if(!vis[i]) dfs(i, 0);
    memset(vis, 0, sizeof(vis));
    for(int i = 1; i <= n; i++)
        if(!vis[i]) getTo(i, 0);
    for(int i = n - 1; i >= 1; i--)
        to[i] = min(to[i], to[i + 1]);
    scanf("%d", &q);
    for(int i = 1; i <= q; i++) {
        int L, R; scanf("%d%d", &L, &R);
        qus[L].push_back(mk(R, i));
    }
    for(int i = n; i >= 1; i--) {
        update(i, to[i], 1, 1, n, 1);
        for(auto& Q : qus[i]) ans[Q.se] = query(i, Q.fi, 1, n, 1);
    }
    for(int i = 1; i <= q; i++) printf("%lld
", ans[i]);
    return 0;
}

/*
*/