AGC002D Stamp Rally [整體二分]

Description:
一張無向聯通圖，給兩個頂點xi$x_i$ ,yi$y_i$ ,問需要最大邊的編號最小使得和兩個點聯通的點數>=zi$>=z_i$ 。

---

Solution:
整體二分，並查集維護連通塊大小。

---

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
struct que {
    int u, v, z, id;
} a[maxn], st1[maxn], st2[maxn];
struct edge {
    int u, v;
} e[maxn];
struct dsu {
    int u, v, szu, szv, fu, fv; 
} st[maxn * 40];
int n, m, top, q;
int fa[maxn], sz[maxn], ans[maxn];
int find(int u) {
    return u == fa[u] ? u : find(fa[u]);
}
bool Y(int u, int v) {
    return find(u) == find(v);
}
void con(int u, int v) {
    if(Y(u, v)) {
        return;
    }
    u = find(u);
    v = find(v);
    if(sz[u] < sz[v]) {
        swap(u, v);
    }
    st[++top] = (dsu){u, v, sz[u], sz[v], fa[u], fa[v]};
    sz[u] += sz[v];
    fa[v] = u;
}
void del(int low) {
    while(top != low) {
        fa[st[top].u] = st[top].fu;
        fa[st[top].v] = st[top].fv;
        sz[st[top].u] = st[top].szu;
        sz[st[top].v] = st[top].szv;
        --top;
    }
}
void cdq(int l, int r, int L, int R) {
    if(L == R || l > r) {
        for(int i = l; i <= r; ++i) {
            ans[a[i].id] = L;
        }
        return;
    }
    int mid = (L + R) >> 1, low = top;
    for(int i = L; i <= mid; ++i) {
        con(e[i].u, e[i].v);
    }
    int p1 = 0, p2 = 0;  
    for(int i = l; i <= r; ++i) {
        if(Y(a[i].u, a[i].v)) {
            if(sz[find(a[i].u)] < a[i].z) {
                st2[++p2] = a[i];
            } else {
                st1[++p1] = a[i];
            }
        } else {
            if(sz[find(a[i].u)] + sz[find(a[i].v)] < a[i].z) {
                st2[++p2] = a[i];
            } else {
                st1[++p1] = a[i];
            }
        }
    } 
    int p = l;
    for(int i = 1; i <= p1; ++i) {
        a[p++] = st1[i];
    }
    for(int i = 1; i <= p2; ++i) {
        a[p++] = st2[i];
    }
    cdq(l + p1, r, mid + 1, R);
    del(low);
    cdq(l, l + p1 - 1, L, mid);
}
int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; ++i) {
        scanf("%d%d", &e[i].u, &e[i].v);
    }
    for(int i = 1; i <= n; ++i) {
        fa[i] = i;
        sz[i] = 1;
    }
    scanf("%d", &q);
    for(int i = 1; i <= q; ++i) {
        scanf("%d%d%d", &a[i].u, &a[i].v, &a[i].z);
        a[i].id = i;
    }
    cdq(1, q, 1, m);
    for(int i = 1; i <= q; ++i) {
        printf("%d\n", ans[i]);
    }
    return 0;
}