Codeforces 1140F Extending Set of Points 线段树 + 按秩合并并查集 (看题解)

Extending Set of Points

我们能发现， 如果把x轴y轴看成点， 那么答案就是在各个连通块里面的x轴的个数乘以y轴的个数之和。

然后就变成了一个并查集的问题， 但是这个题目里面有撤销的操作， 所以我们要把加入和撤销操作变成

这个点影响(L , R)之间的询问， 然后把它丢到线段树里面分成log段， 然后我们dfs一遍线段树， 用按秩合并
并查集取维护， 回溯的时候将并查集撤销。

#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 = 998244353;
const double eps = 1e-6;
const double PI = acos(-1);

int n;
LL ans[N];
map<PII, int> Map;

int fa[N << 1], cntx[N << 1], cnty[N << 1], sz[N << 1];
LL now = 0;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
vector<PII> vc[N << 2];
void Insert(int L, int R, PII e, int l, int r, int rt) {
    if(l >= L && r <= R) {
        vc[rt].push_back(e);
        return;
    }
    int mid = l + r >> 1;
    if(L <= mid) Insert(L, R, e, lson);
    if(R >  mid) Insert(L, R, e, rson);
}

int getRoot(int x) {
    return fa[x] == x ? x : getRoot(fa[x]);
}

void Merge(int x, int y, stack<PII>& stk) {
    int fax = getRoot(x);
    int fay = getRoot(y);
    if(fax != fay) {
        if(sz[fax] < sz[fay]) swap(fax, fay);
        stk.push(mk(fax, fay));
        now -= 1ll * cntx[fax] * cnty[fax];
        now -= 1ll * cntx[fay] * cnty[fay];
        sz[fax] += sz[fay];
        cntx[fax] += cntx[fay];
        cnty[fax] += cnty[fay];
        fa[fay] = fax;
        now += 1ll * cntx[fax] * cnty[fax];
    }
}

void Delete(stack<PII>& stk) {
    while(!stk.empty()) {
        int fax = stk.top().fi;
        int fay = stk.top().se;
        stk.pop();
        now -= 1ll * cntx[fax] * cnty[fax];
        sz[fax] -= sz[fay];
        cntx[fax] -= cntx[fay];
        cnty[fax] -= cnty[fay];
        fa[fay] = fay;
        now += 1ll * cntx[fax] * cnty[fax];
        now += 1ll * cntx[fay] * cnty[fay];
    }
}

void dfs(int l, int r, int rt) {
    stack<PII> stk;
    for(auto& t : vc[rt]) Merge(t.fi, t.se, stk);
    if(l == r) ans[l] = now;
    else {
        int mid = l + r >> 1;
        dfs(l, mid, rt << 1);
        dfs(mid + 1, r, rt << 1 | 1);
    }
    Delete(stk);
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        PII e; scanf("%d%d", &e.fi, &e.se);
        e.se += 300000;
        if(Map.find(e) == Map.end()) {
            Map[e] = i;
        } else {
            Insert(Map[e], i - 1, e, 1, n, 1);
            Map.erase(e);
        }
    }
    for(auto& t : Map) Insert(t.se, n, t.fi, 1, n, 1);
    for(int i = 1; i <= 300000; i++) fa[i] = i, cntx[i] = 1, sz[i] = 1;
    for(int i = 300001; i <= 600000; i++) fa[i] = i, cnty[i] = 1, sz[i] = 1;
    dfs(1, n, 1);
    for(int i = 1; i <= n; i++) printf("%lld ", ans[i]);
    puts("");
    return 0;
 }

/*
*/

原文地址：https://www.cnblogs.com/CJLHY/p/10606629.html