bzoj1018

线段树分治+并查集

线段树本身就是分治结构，碰见这种带删除修改的题目是再合适不过的，我们对于每段修改区间在线段树上打标记，每次路过就进行修改，叶子结点表及答案，先把所有修改在线段树上标记，然后dfs就行了

但是并查集怎么恢复呢？我们只要维护一个栈，保存某次操作之前这两个点的信息，dfs本身就是利用栈来操作，那么每次回溯就用栈的信息恢复之前的样子就行了

#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
int c, tot, cnt, top, ask;
int fa[N], l[N], r[N], d[N], mark[N];
bool ans[N];
struct dsu {
    int u, v, fa, du, dv;
    dsu(int u = 0, int v = 0, int fa = 0, int du = 0, int dv = 0) : u(u), v(v), fa(fa), du(du), dv(dv) {}
} st[N];
pair<int, int> operation[N];
map<pair<int, int>, int> mp;
vector<pair<int, int> > tree[N << 2], q[N << 2];
char opt[10];
inline int read()
{
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}
int find(int x)
{
    return x == fa[x] ? x : find(fa[x]);
}
int id(int x, int y)
{
    return (x - 1) * c + y;
}
void unite(pair<int, int> o)
{
    int x = find(o.first), y = find(o.second);
    if(d[x] > d[y]) swap(x, y);
    st[++top] = dsu(x, y, fa[x], d[x], d[y]);
    if(x != y) 
    {
        d[y] += d[x];
        fa[x] = y;
    }
}
void del(int now)
{
    while(top != now)
    {
        dsu x = st[top];
        fa[x.u] = x.fa;
        d[x.u] = x.du;
        d[x.v] = x.dv;
        --top;
    }
}
void update(int l, int r, int x, int a, int b, pair<int, int> o)
{
    if(l > b || r < a) return;
    if(l >= a && r <= b)
    {
        tree[x].push_back(o);
        return;
    }
    int mid = (l + r) >> 1;
    update(l, mid, x << 1, a, b, o);
    update(mid + 1, r, x << 1 | 1, a, b, o);
}
void dfs(int l, int r, int x)
{
    int now = top;
    for(int i = 0; i < tree[x].size(); ++i) unite(tree[x][i]);
    if(l == r)
    {
        for(int i = 0; i < q[l].size(); ++i)
        {
            pair<int, int> o = q[l][i];
            if(find(o.first) == find(o.second)) puts("Y");
            else puts("N");
        }
        del(now);
        return;
    }
    int mid = (l + r) >> 1;
    dfs(l, mid, x << 1);
    dfs(mid + 1, r, x << 1 | 1);
    del(now); 
}
int main()
{
//    freopen("bzoj_1018.in", "r", stdin);
//    freopen("bzoj_1018.out", "w", stdout);
    memset(l, 0x3f3f, sizeof(l));
    c = read();
    for(int i = 1; i <= 2 * c; ++i)
    {
        fa[i] = i;
        d[i] = 1;
    }
    while(scanf("%s", opt))
    {
        if(opt[0] == 'E') break;
        ++tot;
        int r1 = read(), c1 = read(), r2 = read(), c2 = read(), u = id(r1, c1), v = id(r2, c2);
        if(u > v) swap(u, v);
        if(opt[0] == 'O')
        {        
            if(mp.find(make_pair(u, v)) != mp.end()) continue;
            l[++cnt] = tot;
            operation[cnt] = make_pair(u, v);
            mp[make_pair(u, v)] = cnt;
        }
        if(opt[0] == 'C') 
        {
            if(mp.find(make_pair(u, v)) == mp.end()) continue;
            r[mp[make_pair(u, v)]] = tot - 1;
            mp.erase(make_pair(u, v));
        }
        if(opt[0] == 'A') q[tot].push_back(make_pair(u, v));
    }
    for(int i = 1; i <= cnt; ++i) 
    {
        if(!r[i]) r[i] = tot;
        update(1, tot, 1, l[i], r[i], operation[i]);
    }
    dfs(1, tot, 1);
    for(int i = 1; i <= ask; ++i) if(ans[i]) puts("Y");
    else puts("N");
//    fclose(stdin);
//    fclose(stdout);
    return 0;
}