20200728 SCOI模拟T1（线段树上二分）

T1 P6619 [省选联考 2020 A/B 卷] 冰火战士

思路：
发现选择一个温度时，冰系的贡献时前缀和，火系的贡献是后缀和，答案为较小的贡献的两倍
因为一个贡献单增，一个单减，肯定存在一个点双方贡献相等
因为温度属于整数，所以最佳温度为这个点左右的最近温度
对于一次询问，可以两次二分找到这两个温度
因为要求最大的最佳温度，当零点右侧的点最佳时，可能存在更加右侧的点，最终结果一样，于是需要第三次二分
维护前缀后缀和，可以用线段树
时间复杂度：$O(nlo{g}^{2}n)$，不能通过
看到线段树和二分，考虑线段树上二分，优化掉一个 $logn$
注意卡常

代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include 
using namespace std;
namespace IO {
char _buf[1 << 21], *_p1 = _buf, *_p2 = _buf;
#define ch()                                                                 \
  (_p1 == _p2 &&                                                             \
           (_p2 = (_p1 = _buf) + fread(_buf, 1, 1 << 21, stdin), _p1 == _p2) \
       ? EOF                                                                 \
       : *_p1++)
inline int in() {
  int s = 0, f = 1;
  char x = ch();
  while (x < '0' || x > '9') {
    if (x == '-') f = -1;
    x = ch();
  }
  while (x >= '0' && x <= '9') {
    s = (s * 10) + (x & 15);
    x = ch();
  }
  return f == 1 ? s : -s;
}
char _buf_[1 << 21];
int _p1_ = -1;
inline void flush() {
  fwrite(_buf_, 1, _p1_ + 1, stdout);
  _p1_ = -1;
  return;
}
inline void pc(char x) {
  if (_p1_ == (1 << 21) - 1) flush();
  _buf_[++_p1_] = x;
  return;
}
inline void out(int x) {
  if (!x) pc('0');
  char k[20];
  int tot = 0;
  if (x < 0) {
    pc('-');
    x = -x;
  }
  while (x) {
    k[++tot] = (x % 10) | 48;
    x /= 10;
  }
  for (int i = tot; i; i--) pc(k[i]);
  return;
}
inline void out(string x) {
  int len = x.size();
  for (int i = 0; i < len; i++) pc(x[i]);
  return;
}
}  // namespace IO
using namespace IO;

const int A = 3e6 + 5;
int n;
struct People {
  int id, be, pos, val;
  People(int u = 0, int v = 0, int w = 0, int t = 0) {
    id = u, be = v, pos = w, val = t;
  }
} p[A];
int tmp[A], tot, len;

inline void scan() {
  n = in();
  for (int i = 1; i <= n; i++) {
    int opt = in();
    if (opt == 1) {
      int u = in(), v = in(), w = in();
      p[i] = People(i, u, v, w);
      tmp[++tot] = v;
    } else {
      int u = in();
      p[i] = People(u, -1, 0, 0);
    }
  }
  return;
}

inline int lower(int l, int r, int val) {
  int ans = 0;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (tmp[mid] >= val)
      r = mid - 1, ans = mid;
    else
      l = mid + 1;
  }
  return ans;
}

inline void lsh() {
  sort(tmp + 1, tmp + 1 + tot);
  len = unique(tmp + 1, tmp + 1 + tot) - (tmp + 1);
  for (int i = 1; i <= n; i++)
    if (p[i].be != -1) p[i].pos = lower(1, len, p[i].pos);
  return;
}

struct SGT {
  int l, r, max[2], tag[2];
} tr[4 * A];
int sum[A][2];

#define lch p << 1
#define rch p << 1 | 1
inline void pushup(int p) {
  tr[p].max[0] = max(tr[lch].max[0], tr[rch].max[0]);
  tr[p].max[1] = max(tr[lch].max[1], tr[rch].max[1]);
  return;
}
inline void pushdown(int p) {
  if (tr[p].l != tr[p].r && (tr[p].tag[0] || tr[p].tag[1])) {
    if (tr[p].tag[0]) {
      tr[lch].max[0] += tr[p].tag[0];
      tr[lch].tag[0] += tr[p].tag[0];
      tr[rch].max[0] += tr[p].tag[0];
      tr[rch].tag[0] += tr[p].tag[0];
      tr[p].tag[0] = 0;
    }
    if (tr[p].tag[1]) {
      tr[lch].max[1] += tr[p].tag[1];
      tr[lch].tag[1] += tr[p].tag[1];
      tr[rch].max[1] += tr[p].tag[1];
      tr[rch].tag[1] += tr[p].tag[1];
      tr[p].tag[1] = 0;
    }
  }
  return;
}
inline void build(int p, int l, int r) {
  tr[p].l = l, tr[p].r = r;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build(lch, l, mid), build(rch, mid + 1, r);
  return;
}
inline void add(int p, int l, int r, int val, int pos) {
  if (tr[p].l >= l && tr[p].r <= r) {
    tr[p].max[pos] += val;
    tr[p].tag[pos] += val;
    return;
  }
  pushdown(p);
  int mid = (tr[p].l + tr[p].r) >> 1;
  if (l <= mid) add(lch, l, r, val, pos);
  if (r >= mid + 1) add(rch, l, r, val, pos);
  pushup(p);
  return;
}
inline pair<int, int> qurey1(int p) {
  if (tr[p].l == tr[p].r)
    return make_pair(min(tr[p].max[0], tr[p].max[1]) * 2, tr[p].l);
  pushdown(p);
  if ((tr[lch].max[0] + sum[tr[rch].l][0]) - tr[rch].max[1] > 0)
    return qurey1(lch);
  else
    return qurey1(rch);
}
inline pair<int, int> qurey2(int p) {
  if (tr[p].l == tr[p].r)
    return make_pair(min(tr[p].max[0], tr[p].max[1]) * 2, tr[p].l);
  pushdown(p);
  if (tr[lch].max[0] - (tr[rch].max[1] + sum[tr[lch].r][1]) < 0)
    return qurey2(rch);
  else
    return qurey2(lch);
}
inline int qurey3(int p, int l, int val) {
  if (tr[p].l == tr[p].r) return tr[p].l;
  pushdown(p);
  int mid = (tr[p].l + tr[p].r) >> 1;
  if (mid < l) return qurey3(rch, l, val);
  if (min(tr[rch].max[1], tr[lch].max[0] + sum[tr[rch].l][0]) * 2 == val)
    return qurey3(rch, l, val);
  else
    return qurey3(lch, l, val);
}
#undef lch
#undef rch

inline void solve() {
  pair<int, int> p1 = qurey1(1);
  pair<int, int> p2 = qurey2(1);
  if (!p1.first && !p2.first) {
    out("Peace\n");
    return;
  }
  if (p1.first > p2.first) {
    out(tmp[p1.second]), pc(' '), out(p1.first), pc('\n');
    return;
  } else {
    int w = qurey3(1, p2.second, p2.first);
    out(tmp[w]), pc(' '), out(p2.first), pc('\n');
  }
  return;
}

inline void work(int now) {
  if (p[now].be == -1) {
    int x = p[now].id;
    if (!p[x].be) {
      add(1, p[x].pos, len, -p[x].val, 0);
      sum[p[x].pos][0] -= p[x].val;
    } else {
      add(1, 1, p[x].pos, -p[x].val, 1);
      sum[p[x].pos][1] -= p[x].val;
    }
  } else {
    if (!p[now].be) {
      add(1, p[now].pos, len, p[now].val, 0);
      sum[p[now].pos][0] += p[now].val;
    } else {
      add(1, 1, p[now].pos, p[now].val, 1);
      sum[p[now].pos][1] += p[now].val;
    }
  }
  solve();
  return;
}

signed main() {
  // freopen("icefire.in", "r", stdin);
  // freopen("icefire.out", "w", stdout);
  scan();
  lsh();
  build(1, 1, len);
  for (int i = 1; i <= n; i++) work(i);
  IO::flush();
  return 0;
}