什么是矢？这就是矢

Easy problem I HDU - 7284

思路

没有任何思维，如果你想简单点做这个题，那么你还是止步吧，这个题目直接用最简单的思路，分类讨论，按照绝对值内正负分别建立线段树，由于 x 不减，所以从一个树跳到另一个树最多一次，总时间复杂度 \(O((n + m)logn)\)

由于思路太短了，所以我放个图凑个数（

code

代码长的吓死你

#include <iostream>
#define int long long

using namespace std;

const int MaxN = 2e5 + 10;

int a[MaxN], n, m, t;

namespace seg2 {
int d[MaxN << 2], tag1[MaxN << 2], tag2[MaxN << 2], len[MaxN << 2];

void build(int l = 1, int r = n, int p = 1) {
  tag1[p] = tag2[p] = 0;
  if (l == r) {
    d[p] = len[p] = 0;
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, p << 1), build(mid + 1, r, p << 1 | 1);
  d[p] = d[p << 1] + d[p << 1 | 1], len[p] = len[p << 1] + len[p << 1 | 1];
}

void addtag1(int p) {
  if (len[p]) d[p] = -d[p], tag2[p] = -tag2[p], tag1[p] ^= 1;
}

void addtag2(int p, int w) {
  if (len[p]) d[p] += len[p] * w, tag2[p] += w;
}

void pd(int p) {
  if (tag1[p]) addtag1(p << 1), addtag1(p << 1 | 1), tag1[p] = 0;
  if (tag2[p]) addtag2(p << 1, tag2[p]), addtag2(p << 1 | 1, tag2[p]), tag2[p] = 0;
}

void update(int p) {
  d[p] = d[p << 1] + d[p << 1 | 1], len[p] = len[p << 1] + len[p << 1 | 1];
}

void modify(int pl, int pr, int w, int l = 1, int r = n, int p = 1) {
  if (!len[p]) return;
  if (pl <= l && r <= pr) {
    return addtag1(p), addtag2(p, w);
  }
  if (pl > r || pr < l) return;
  pd(p);
  int mid = l + r >> 1;
  modify(pl, pr, w, l, mid, p << 1), modify(pl, pr, w, mid + 1, r, p << 1 | 1), update(p);
}

void mao(int k, int w, int l = 1, int r = n, int p = 1) {
  if (l == r) {
    d[p] = w, len[p] = 1;
    return;
  }
  pd(p);
  int mid = l + r >> 1;
  if (k <= mid) mao(k, w, l, mid, p << 1);
  if (k > mid) mao(k, w, mid + 1, r, p << 1 | 1);
  update(p);
}

int query(int pl, int pr, int l = 1, int r = n, int p = 1) {
  if (!len[p]) return 0;
  if (pl <= l && r <= pr) return d[p];
  if (pl > r || pr < l) return 0;
  pd(p);
  int mid = l + r >> 1;
  return query(pl, pr, l, mid, p << 1) + query(pl, pr, mid + 1, r, p << 1 | 1);
}
};

namespace seg1 {
int d[MaxN << 2], tag[MaxN << 2], mix[MaxN << 2], len[MaxN << 2];

void build(int l = 1, int r = n, int p = 1) {
  tag[p] = 0;
  if (l == r) {
    d[p] = mix[p] = a[l], len[p] = 1;
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, p << 1), build(mid + 1, r, p << 1 | 1);
  d[p] = d[p << 1] + d[p << 1 | 1], mix[p] = min(mix[p << 1], mix[p << 1 | 1]), len[p] = len[p << 1] + len[p << 1 | 1];
}

void addtag(int p, int w) {
  if (len[p]) d[p] -= w * len[p], mix[p] -= w, tag[p] += w;
}

void pd(int p) {
  addtag(p << 1, tag[p]), addtag(p << 1 | 1, tag[p]), tag[p] = 0;
}

void update(int p) {
  d[p] = d[p << 1] + d[p << 1 | 1], mix[p] = min(mix[p << 1], mix[p << 1 | 1]), len[p] = len[p << 1] + len[p << 1 | 1];
}

void modify(int pl, int pr, int w, int l = 1, int r = n, int p = 1) {
  if (!len[p]) return;
  if (pl > r || pr < l) return;
  if (l == r && mix[p] <= w) {
    len[p] = d[p] = 0;
    return seg2::mao(l, mix[p]), void(mix[p] = 1e18);
  }
  if (pl <= l && r <= pr) {
    if (mix[p] <= w) {
      pd(p);
      int mid = l + r >> 1;
      return modify(pl, pr, w, l, mid, p << 1), modify(pl, pr, w, mid + 1, r, p << 1 | 1), update(p);
    }
    return addtag(p, w);
  }
  pd(p);
  int mid = l + r >> 1;
  modify(pl, pr, w, l, mid, p << 1), modify(pl, pr, w, mid + 1, r, p << 1 | 1), update(p);
}

int query(int pl, int pr, int l = 1, int r = n, int p = 1) {
  if (!len[p]) return 0;
  if (pl <= l && r <= pr) return d[p];
  if (pl > r || pr < l) return 0;
  pd(p);
  int mid = l + r >> 1;
  return query(pl, pr, l, mid, p << 1) + query(pl, pr, mid + 1, r, p << 1 | 1);
}
};  // namespace seg1

signed main() {
  ios::sync_with_stdio(0), cin.tie(0);
  for (cin >> t; t; t--) {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }    
    seg1::build(), seg2::build();
    for (int op, l, r, x; m; m--) {
      cin >> op >> l >> r;
      if (op == 1) {
        cin >> x;
        seg1::modify(l, r, x);
        seg2::modify(l, r, x);
      } else {
        cout << seg1::query(l, r) + seg2::query(l, r) << '\n';
      }
    }
  }
  return 0;
}