线段树分治初步学习&洛谷P5227[AHOI2013]连通图

线段树分治

其实思想说起来是比较简单的，我们把这个题里的所有操作（比如连边删边查询balabala）全部拍到一棵线段树上，然后对着整棵树dfs一下求解答案，顺便把操作做一下，回溯的时候撤销一下即可。虽然有的操作需要以区间形式拍到树上，导致它可能会被拆成两个，但线段树的形态同样保证了操作最多只会被拆分\(log(区间长度)\)次，保障了复杂度。

洛谷P5227[AHOI2013]连通图

其实就是线段树分治+带撤销并查集，并查集写按秩合并，不能路径压缩（否则会破坏结构，就会撤销出奇怪的效果）
以询问的时间轴为下标建一棵线段树，然后把边存在的时间区间拍到线段树上，用\(vector\)存下来，再对着树一波\(dfs\)记录答案，注意在叶节点不能写\(return\)我沙茶了，不然叶节点如果有操作就会还没撤销就返回了。

/*P5227 [AHOI2013]连通图*/  
#include <bits/stdc++.h>
#define N (100000 + 5)
using namespace std;
inline int read() {
    int cnt = 0, f = 1; char c = getchar();
    while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
    while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();}
    return cnt * f;
}
int n, m, k, c, x, top;
int fa[N];
struct node {
    int u, v;
}edge[N << 1];
struct node2 {
    int siz, dep, fa;
    node2(int siz_ = 1, int dep_ = 1, int fa_ = 0) : siz(siz_), dep(dep_), fa(fa_){};
}bcj[N << 1], ctrl_Z[N << 1];
int cur[N << 1], top2;
int pre[N << 1];
int get_father(int x) {return x == bcj[x].fa ? x : get_father(bcj[x].fa);}
void merge(int p, int q) {
    int x = get_father(p), y = get_father(q);
    if (x == y) return;
    if (bcj[x].dep > bcj[y].dep) swap(x, y);
    ctrl_Z[++top] = bcj[x];
    ctrl_Z[++top] = bcj[y];
    bcj[x].fa = y, bcj[y].dep = max(bcj[y].dep, bcj[x].dep + 1), bcj[y].siz += bcj[x].siz;
    cur[++top2] = x;
    cur[++top2] = y;
}
struct node3{
    int l, r;
    vector<node> E;
    #define l(p) tree[p].l
    #define r(p) tree[p].r
}tree[N << 2];

void build(int l, int r, int p) {
    l(p) = l, r(p) = r;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build (l, mid, p << 1);
    build (mid + 1, r, p << 1 | 1);
}

void insert(node x, int l, int r, int p) {
    if (l <= l(p) && r >= r(p)) { tree[p].E.push_back(x); return; }
    register int mid = (l(p) + r(p)) >> 1;
    if (l <= mid) insert(x, l, r, p << 1);
    if (r > mid) insert(x, l, r, p << 1 | 1);
}
void dfs_(int p) {
    int tp = top2;
    for (register unsigned int i = 0; i < tree[p].E.size(); i++) {
        node now = tree[p].E[i];
        merge(now.u, now.v);
    }
    if (l(p) == r(p)) {
        int now = bcj[get_father(1)].siz;
        printf(now == n ? "Connected\n" : "Disconnected\n");
    } else dfs_(p << 1), dfs_(p << 1 | 1);
    for (; top2 > tp; --top2, --top) {bcj[cur[top2]] = ctrl_Z[top];}
}

int main() {
    n = read(), m = read();
    for (register int i = 1; i <= n; i++) bcj[i] = node2(1, 1, i);
    for (register int i = 1; i <= m; i++) 
        edge[i].u = read(), edge[i].v = read(), pre[i] = 1;
    k = read();
    build (1, k, 1);
    for (register int i = 1; i <= k; i++) {
        c = read();
        for (register int j = 1; j <= c; j++) {
            x = read();
            if (pre[x] < i) insert(edge[x], pre[x], i - 1, 1);
            pre[x] = i + 1;
        }
    }
    for (register int i = 1; i <= m; i++) {
        if (pre[i] <= k) insert(edge[i], pre[i], k, 1);
    }
    dfs_(1);
    return 0;
} 

标签：dep,洛谷,bcj,int,线段,top,AHOI2013,++,P5227
来源： https://www.cnblogs.com/kma093/p/11275762.html