洛谷 P2147 [SDOI2008]洞穴勘测 (线段树分治)

题目链接

题解

早就想写线段树分治的题了。

对于每条边，它存在于一段时间

我们按时间来搞

我们可把一条边看做一条线段

我们可以模拟线段树操作，不断分治下去

把覆盖\(l-r\)这段时间的线段筛选出来,用并查集维护联通性，回溯时撤销操作

注意不能使用路径压缩（不能破坏树的结构，方便撤销操作）

Code

#include<bits/stdc++.h>
#define LL long long
#define RG register
using namespace std;

inline int gi() {
    int f = 1, s = 0;
    char c = getchar();
    while (c != '-' && (c < '0' || c > '9')) c = getchar();
    if (c == '-') f = -1, c = getchar();
    while (c >= '0' && c <= '9') s = s*10+c-'0', c = getchar();
    return f == 1 ? s : -s;
}

const int N = 10010, M = 200010;

struct question {
    int time, u, v;
}q[M];
int ql, w;

struct bian {
    int u, v, s, e;
};
vector<bian> g;
map<pair<int, int>, int> Mp;
int siz[N], fa[N];
inline int find(int x) {
    return x == fa[x] ? x : find(fa[x]);
}
pair<int, int> stk[M];
int top;

void link(int x, int y) {//按秩合并
    x = find(x); y = find(y);
    if (x == y) {
        stk[++top] = make_pair(0, 0);
        return ;
    }
    if (siz[x] < siz[y]) swap(x, y);
    stk[++top] = make_pair(x, y);//y接在x上
    fa[y] = x;
    siz[x] += siz[y];
    return ;
}

void clear() {//撤销操作
    int x = stk[top].first, y = stk[top--].second;
    if (!x && !y) return ;
    fa[y] = y;
    siz[x] -= siz[y];
    return ;
}

void divide(int l, int r, vector<bian> E) {
    vector<bian> L, R;
    int tmp = top, mid = (l + r) >> 1;
    for (int i = 0; i < (int)E.size(); i++) {
        if (E[i].s <= l && E[i].e >= r) link(E[i].u, E[i].v);
        else {
            if (E[i].s <= mid) L.push_back(E[i]);
            if (E[i].e > mid) R.push_back(E[i]);
        }
    }
    if (l == r) {
        while (q[w].time == l && w <= ql) {
            if (find(q[w].u) == find(q[w].v)) printf("Yes\n");
            else printf("No\n");
            w++;
        }
        if (w > ql) exit(0);
        return ;
    }
    else divide(l, mid, L), divide(mid+1, r, R);
    while (top > tmp) clear();
    return ;
}

int main() {
    int n = gi(), m = gi(), x, y;
    char s[10];
    for (int i = 1; i <= m; i++) {
        cin >> s >> x >> y;
        if (x > y) swap(x, y);
        if (s[0] == 'C') {
            if (Mp.find(make_pair(x, y)) == Mp.end())
                g.push_back((bian){x, y, i, m}), Mp[make_pair(x, y)] = g.size()-1;
        }
        else if (s[0] == 'D') {
            g[Mp[make_pair(x, y)]].e = i-1;
            Mp.erase(Mp.find(make_pair(x, y)));
        }
        else q[++ql] = (question) {i, x, y};
    }
    /*for (int i = 0; i < g.size(); i++)
        printf("%d %d %d %d\n", g[i].u, g[i].v, g[i].s, g[i].e);*/
    w = 1;
    for (int i = 1; i <= n; i++) fa[i] = i, siz[i] = 1;
    divide(1, m, g);
    return 0;
}