• P5163 WD与地图 [整体二分,强连通分量,线段树合并]


    首先不用说,倒着操作。整体二分来做强连通分量,然后线段树合并,这题就做完了。

    // powered by c++11
    // by Isaunoya
    #include <bits/stdc++.h>
    #define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
    #define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
    using namespace std;
    using db = double;
    using ll = long long;
    using uint = unsigned int;
    #define int long long
    using pii = pair<int, int>;
    #define ve vector
    #define Tp template
    #define all(v) v.begin(), v.end()
    #define sz(v) ((int)v.size())
    #define pb emplace_back
    #define fir first
    #define sec second
    // the cmin && cmax
    Tp<class T> void cmax(T& x, const T& y) {
      if (x < y) x = y;
    }
    Tp<class T> void cmin(T& x, const T& y) {
      if (x > y) x = y;
    }
    // sort , unique , reverse
    Tp<class T> void sort(ve<T>& v) { sort(all(v)); }
    Tp<class T> void unique(ve<T>& v) {
      sort(all(v));
      v.erase(unique(all(v)), v.end());
    }
    Tp<class T> void reverse(ve<T>& v) { reverse(all(v)); }
    const int SZ = 0x191981;
    struct FILEIN {
      ~FILEIN() {}
      char qwq[SZ], *S = qwq, *T = qwq, ch;
      char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
      FILEIN& operator>>(char& c) {
        while (isspace(c = GETC()))
          ;
        return *this;
      }
      FILEIN& operator>>(string& s) {
        while (isspace(ch = GETC()))
          ;
        s = ch;
        while (!isspace(ch = GETC())) s += ch;
        return *this;
      }
      Tp<class T> void read(T& x) {
        bool sign = 1;
        while ((ch = GETC()) < 0x30)
          if (ch == 0x2d) sign = 0;
        x = (ch ^ 0x30);
        while ((ch = GETC()) > 0x2f) x = x * 0xa + (ch ^ 0x30);
        x = sign ? x : -x;
      }
      FILEIN& operator>>(int& x) { return read(x), *this; }
      FILEIN& operator>>(signed& x) { return read(x), *this; }
      FILEIN& operator>>(unsigned& x) { return read(x), *this; }
    } in;
    struct FILEOUT {
      const static int LIMIT = 0x114514;
      char quq[SZ], ST[0x114];
      signed sz, O;
      ~FILEOUT() { sz = O = 0; }
      void flush() {
        fwrite(quq, 1, O, stdout);
        fflush(stdout);
        O = 0;
      }
      FILEOUT& operator<<(char c) { return quq[O++] = c, *this; }
      FILEOUT& operator<<(string str) {
        if (O > LIMIT) flush();
        for (char c : str) quq[O++] = c;
        return *this;
      }
      Tp<class T> void write(T x) {
        if (O > LIMIT) flush();
        if (x < 0) {
          quq[O++] = 0x2d;
          x = -x;
        }
        do {
          ST[++sz] = x % 0xa ^ 0x30;
          x /= 0xa;
        } while (x);
        while (sz) quq[O++] = ST[sz--];
        return;
      }
      FILEOUT& operator<<(int x) { return write(x), *this; }
      FILEOUT& operator<<(signed x) { return write(x), *this; }
      FILEOUT& operator<<(unsigned x) { return write(x), *this; }
    } out;
    
    int n, m, qwq;
    const int maxn = 1e6 + 61;
    const int lim = 1e9;
    struct edge {
      int v, nxt;
    } e[maxn];
    int head[maxn], tot = 0;
    struct node {
      int u, v, t;
    } a[maxn], b[maxn];
    vector<node> path[maxn];
    int rt[maxn], ls[maxn << 5], rs[maxn << 5], s[maxn << 5], sum[maxn << 5];
    int fa[maxn], low[maxn], dfn[maxn], st[maxn], ins[maxn], q[maxn];
    int type[maxn], aa[maxn], bb[maxn], val[maxn], cnt = 0, top = 0, h = 0, idx = 0;
    set<pii> se;
    int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
    void unite(int x, int y) {
      if ((x = find(x)) ^ (y = find(y))) fa[x] = y;
    }
    void tarjan(int u) {
      dfn[u] = low[u] = ++idx, st[++top] = u, ins[u] = 1;
      for (int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].v;
        if (!dfn[v]) {
          tarjan(v);
          cmin(low[u], low[v]);
        } else if (ins[v])
          cmin(low[u], dfn[v]);
      }
      if (low[u] == dfn[u]) do {
          ins[st[top]] = 0, unite(st[top], u);
        } while (st[top--] != u);
    }
    void upd(int& p, int l, int r, int x, int v) {
      if (!p) p = ++cnt;
      s[p] += v, sum[p] += v * x;
      if (l == r) return;
      int mid = l + r >> 1;
      (x <= mid) ? upd(ls[p], l, mid, x, v) : upd(rs[p], mid + 1, r, x, v);
    }
    int qry(int p, int l, int r, int k) {
      if (k >= s[p]) return sum[p];
      if (l == r) return l * k;
      int mid = l + r >> 1, x = s[rs[p]];
      if (x >= k)
        return qry(rs[p], mid + 1, r, k);
      else
        return sum[rs[p]] + qry(ls[p], l, mid, k - x);
    }
    int merge(int x, int y) {
      if (!x || !y) return x | y;
      sum[x] += sum[y], s[x] += s[y];
      ls[x] = merge(ls[x], ls[y]), rs[x] = merge(rs[x], rs[y]);
      return x;
    }
    void solve(int l, int r, int ql, int qr) {
      if (l == r) {
        for (int i = ql; i <= qr; i++) path[l].push_back(a[i]);
        return;
      }
      if (ql == qr) {
        if (find(a[ql].u) == find(a[ql].v)) path[a[ql].t].push_back(a[ql]);
        return;
      }
      int mid = l + r >> 1;
      tot = h = 0;
      for (int i = ql; i <= qr; i++)
        if (a[i].t <= mid) {
          int u = find(a[i].u), v = find(a[i].v);
          q[++h] = u, q[++h] = v;
          head[u] = head[v] = 0;
        }
      for (int i = 1; i <= h; i += 2) {
        int u = q[i], v = q[i + 1];
        e[++tot] = { v, head[u] }, head[u] = tot;
        dfn[u] = low[u] = ins[u] = 0;
        dfn[v] = low[v] = ins[v] = 0;
      }
      top = idx = 0;
      for (int i = 1; i <= h; i++)
        if (!dfn[q[i]]) tarjan(q[i]);
      vector<pii> mp;
      int t1 = 0, t2 = ql;
      h = 1;
      for (int i = ql; i <= qr; i++) {
        int flag = 0;
        if (a[i].t <= mid) {
          int u = q[h], v = q[h + 1];
          h += 2;
          if (find(u) == find(v)) flag = 1;
          mp.push_back({ u, fa[u] }), mp.push_back({ v, fa[v] });
        }
        if (flag)
          a[t2++] = a[i];
        else
          b[++t1] = a[i];
      }
      for (int i = t2; i <= qr; i++) a[i] = b[i - t2 + 1];
      h--;
      for (int i = 1; i <= h; i++) fa[q[i]] = q[i];
      if (ql < t2) solve(l, mid, ql, t2 - 1);
      for (auto x : mp) fa[x.first] = x.second;
      if (qr >= t2) solve(mid + 1, r, t2, qr);
    }
    signed main() {
    #ifdef _WIN64
      freopen("testdata.in", "r", stdin);
    #else
      ios_base ::sync_with_stdio(false);
      cin.tie(nullptr), cout.tie(nullptr);
    #endif
      // code begin.
      in >> n >> m >> qwq;
      for (int i = 1; i <= n; i++) in >> val[i], fa[i] = i;
      for (int i = 1; i <= m; i++) {
        int u, v;
        in >> u >> v, se.insert({ u, v });
      }
      for (int i = qwq; i; i--) {
        int op, x, y;
        in >> op >> x >> y;
        type[i] = op, aa[i] = x, bb[i] = y;
        if (op == 1) a[++tot] = { x, y, i }, se.erase({ x, y });
        if (op == 2) val[x] += y;
      }
      for (pii x : se) a[++tot] = { x.first, x.second, 0 };
      solve(0, qwq + 1, 1, m), tot = top = 0, type[0] = 1;
      for (int i = 1; i <= n; i++) fa[i] = i;
      for (int i = 1; i <= n; i++) upd(rt[i], 1, lim, val[i], 1);
      vector<int> ans;
      for (int i = 0; i <= qwq; i++) {
        if (type[i] == 1) {
          for (auto x : path[i]) {
            x.u = find(x.u), x.v = find(x.v);
            if (x.u == x.v) continue;
            fa[x.v] = x.u, rt[x.u] = merge(rt[x.u], rt[x.v]);
          }
        }
        if (type[i] == 2) {
          int u = find(aa[i]);
          upd(rt[u], 1, lim, val[aa[i]], -1);
          val[aa[i]] -= bb[i];
          upd(rt[u], 1, lim, val[aa[i]], 1);
        }
        if (type[i] == 3) {
          int u = find(aa[i]);
          ans.push_back(qry(rt[u], 1, lim, bb[i]));
        }
      }
      reverse(ans);
      for (int x : ans) out << x << '
    ';
      return out.flush(), 0;
      // code end.
    }
    
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  • 原文地址:https://www.cnblogs.com/Isaunoya/p/12285263.html
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