Codeforces 466E Information Graph

Information Graph

把询问离线之后就能随便搞了， 去check一下是不是祖先， 可以用倍增也能用dfs序。

#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 = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;

int n, m, cnt, deg[N], fa[N], op[N], x[N], y[N], depth[N], f[N][20];
bool vis[N];
bool ans[N];
vector<int> G[N];
vector<int> qus[N];

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

void Merge(int x, int y) {
    fa[getRoot(x)] = getRoot(y);
}

void dfs(int u, int fa, int idx) {
    vis[u] = idx;
    depth[u] = depth[fa] + 1;
    f[u][0] = fa;
    for(int i = 1; i < 20; i++)
        f[u][i] = f[f[u][i - 1]][i - 1];
    for(int& v : G[u]) if(!vis[v]) dfs(v, u, idx);
}

bool check(int u, int v) {
    if(depth[u] < depth[v]) return false;
    if(vis[u] != vis[v]) return false;
    if(getRoot(u) != getRoot(v)) return false;
    for(int i = 19; i >= 0; i--)
        if(depth[f[u][i]] >= depth[v])
            u = f[u][i];
    return u == v;
}

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) fa[i] = i;
    for(int i = 1; i <= m; i++) {
        scanf("%d%d", &op[i], &x[i]);
        if(op[i] != 2) scanf("%d", &y[i]);
        if(op[i] == 3) qus[y[i]].push_back(i);
        if(op[i] == 1) G[y[i]].push_back(x[i]), deg[x[i]]++;
    }
    for(int i = 1; i <= n; i++)
        if(!deg[i]) dfs(i, 0, ++cnt);
    cnt = 0;
    for(int i = 1; i <= m; i++) {
        if(op[i] == 1) {
            Merge(x[i], y[i]);
        } else if(op[i] == 2) {
            cnt++;
            for(auto& p : qus[cnt])
                if(check(x[i], x[p]))
                    ans[p] = true;
        }
    }
    for(int i = 1; i <= m; i++)
        if(op[i] == 3) printf("%s
", ans[i] ? "YES" : "NO");
    return 0;
}

/**/