• [bzoj3226][Sdoi2008]校门外的区间——线段树


    题目

    题解

    直接套黄学长模板。
    Orz

    代码

    #include <bits/stdc++.h>
    using namespace std;
    #define ll long long
    #define inf 1000000000
    #define n (65536 * 2 + 1)
    char ch[5];
    int read() {
      int x = 0, f = 0;
      char ch = getchar();
      while (ch < '0' || ch > '9') {
        if (ch == '(')
          f = -1;
        ch = getchar();
      }
      while (ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
      }
      if (ch == ')')
        f = 1;
      return x * 2 - f;
    }
    struct seg {
      int l, r, val, tag, rev;
    } t[4 * n];
    
    void build(int k, int l, int r) {
      t[k].l = l;
      t[k].r = r;
      t[k].tag = -1;
      if (l == r)
        return;
      int mid = (l + r) >> 1;
      build(k << 1, l, mid);
      build(k << 1 | 1, mid + 1, r);
    }
    
    void pushdown(int k) {
      int tag = t[k].tag, rev = t[k].rev;
      t[k].tag = -1;
      t[k].rev = 0;
      if (t[k].l == t[k].r) {
        if (tag != -1)
          t[k].val = tag;
        t[k].val ^= rev;
        return;
      }
      if (tag != -1) {
        t[k << 1].tag = t[k << 1 | 1].tag = tag;
        t[k << 1].rev = t[k << 1 | 1].rev = 0;
      }
      t[k << 1].rev ^= rev;
      t[k << 1 | 1].rev ^= rev;
    }
    int query(int k, int x) {
      pushdown(k);
      int l = t[k].l, r = t[k].r;
      if (l == r)
        return t[k].val;
      int mid = (l + r) >> 1;
      if (x <= mid)
        return query(k << 1, x);
      else
        return query(k << 1 | 1, x);
    }
    void modify(int k, int x, int y, int val) {
      if (y < x)
        return;
      pushdown(k);
      int l = t[k].l, r = t[k].r;
      if (l == x && r == y) {
        if (val == -1)
          t[k].rev ^= 1;
        else
          t[k].tag = val;
        return;
      }
      int mid = (l + r) >> 1;
      if (y <= mid)
        modify(k << 1, x, y, val);
      else if (x > mid)
        modify(k << 1 | 1, x, y, val);
      else {
        modify(k << 1, x, mid, val);
        modify(k << 1 | 1, mid + 1, y, val);
      }
    }
    void rever(int k, int x, int y) { modify(k, x, y, -1); }
    int main() {
      build(1, 1, n);
      while (scanf("%s", ch) != EOF) {
        int a = read(), b = read();
        a += 2;
        b += 2;
        switch (ch[0]) {
        case 'U':
          modify(1, a, b, 1);
          break;
        case 'I':
          modify(1, 1, a - 1, 0);
          modify(1, b + 1, n, 0);
          break;
        case 'D':
          modify(1, a, b, 0);
          break;
        case 'C':
          modify(1, 1, a - 1, 0);
          modify(1, b + 1, n, 0);
          rever(1, a, b);
          break;
        case 'S':
          rever(1, a, b);
          break;
        }
      }
      int start = -1, last = -1, flag = 0;
      for (int i = 1; i <= n; i++) {
        if (query(1, i)) {
          if (start == -1)
            start = i;
          last = i;
        } else {
          if (start != -1) {
            if (flag)
              printf(" ");
            else
              flag = 1;
            if (start & 1)
              printf("(");
            else
              printf("[");
            printf("%d", start / 2 - 1);
            printf(",");
            printf("%d", (last + 1) / 2 - 1);
            if (last & 1)
              printf(")");
            else
              printf("]");
          }
          last = start = -1;
        }
      }
      if (!flag)
        printf("empty set");
      return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/gengchen/p/6437378.html
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