• [ZJOI2013]K大数查询


    嘟嘟嘟


    这应该算是整体二分入门题吧。


    唯一的区别就是这次是区间修改,树状数组可能不太好做,换成线段树就好了。
    然后每一次别忘了清空。

    #include<cstdio>
    #include<iostream>
    #include<cmath>
    #include<algorithm>
    #include<cstring>
    #include<cstdlib>
    #include<cctype>
    #include<vector>
    #include<stack>
    #include<queue>
    using namespace std;
    #define enter puts("") 
    #define space putchar(' ')
    #define Mem(a, x) memset(a, x, sizeof(a))
    #define rg register
    typedef long long ll;
    typedef double db;
    const int INF = 0x3f3f3f3f;
    const db eps = 1e-8;
    const int maxn = 5e4 + 5;
    inline ll read()
    {
      ll ans = 0;
      char ch = getchar(), last = ' ';
      while(!isdigit(ch)) last = ch, ch = getchar();
      while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
      if(last == '-') ans = -ans;
      return ans;
    }
    inline void write(ll x)
    {
      if(x < 0) x = -x, putchar('-');
      if(x >= 10) write(x / 10);
      putchar(x % 10 + '0');
    }
    
    int n, m, cnt = 0;
    struct Node
    {
      int id, L, R; ll d;
    }t[maxn], tl[maxn], tr[maxn];
    int ans[maxn];
    
    int l[maxn << 2], r[maxn << 2];
    ll sum[maxn << 2], lzy[maxn << 2];
    inline void build(int L, int R, int now)
    {
      l[now] = L; r[now] = R;
      if(L == R) return;
      int mid = (L + R) >> 1;
      build(L, mid, now << 1);
      build(mid + 1, R, now << 1 | 1);
    }
    inline void pushdown(int now)
    {
      if(lzy[now])
        {
          sum[now << 1] += (ll)(r[now << 1] - l[now << 1] + 1) * lzy[now];
          sum[now << 1 | 1] += (ll)(r[now << 1 | 1] - l[now << 1 | 1] + 1) * lzy[now];
          lzy[now << 1] += lzy[now]; lzy[now << 1 | 1] += lzy[now];
          lzy[now] = 0;
        }
    }
    inline void update(int L, int R, int now, int d)
    {
      if(L == l[now] && R == r[now])
        {
          sum[now] += (R - L + 1) * d; lzy[now] += d;
          return;
        }
      pushdown(now);
      int mid = (l[now] + r[now]) >> 1;
      if(R <= mid) update(L, R, now << 1, d);
      else if(L > mid) update(L, R, now << 1 | 1, d);
      else update(L, mid, now << 1, d), update(mid + 1, R, now << 1 | 1, d);
      sum[now] = sum[now << 1] + sum[now << 1 | 1];
    }
    inline ll query(int L, int R, int now)
    {
      if(L == l[now] && R == r[now]) return sum[now];
      pushdown(now);
      int mid = (l[now] + r[now]) >> 1;
      if(R <= mid) return query(L, R, now << 1);
      else if(L > mid) return query(L, R, now << 1 | 1);
      else return query(L, mid, now << 1) + query(mid + 1, R, now << 1 | 1);
    }
    
    inline void solve(int vl, int vr, int ql, int qr)
    {
      if(ql > qr) return;
      if(vl == vr)
        {
          for(int i = ql; i <= qr; ++i)
    	if(t[i].id) ans[t[i].id] = vl;
          return;
        }
      int mid = (vl + vr) >> 1;
      int id1 = 0, id2 = 0;
      for(int i = ql; i <= qr; ++i)
        {
          if(!t[i].id)
    	{
    	  if(t[i].d > mid) update(t[i].L, t[i].R, 1, 1), tr[++id2] = t[i];
    	  else tl[++id1] = t[i];
    	}
          else
    	{
    	  ll tp = query(t[i].L, t[i].R, 1);
    	  if(tp < t[i].d) t[i].d -= tp, tl[++id1] = t[i];
    	  else tr[++id2] = t[i];
    	}
        }
      for(int i = 1; i <= id2; ++i) if(!tr[i].id) update(tr[i].L, tr[i].R, 1, -1);
      for(int i = 1; i <= id1; ++i) t[ql + i - 1] = tl[i];
      for(int i = 1; i <= id2; ++i) t[ql + id1 + i - 1] = tr[i];
      solve(vl, mid, ql, ql + id1 - 1);
      solve(mid + 1, vr, ql + id1, qr);
    }
    
    int main()
    {
      n = read(); m = read();
      build(1, n, 1);
      for(int i = 1; i <= m; ++i)
        {
          int op = read(), L = read(), R = read(), d = read();
          t[i] = (Node){op == 2 ? ++cnt : 0, L, R, d};
        }
      solve(-n, n, 1, m);
      for(int i = 1; i <= cnt; ++i) write(ans[i]), enter;
      return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/mrclr/p/10144022.html
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