HDU 4614 Vases and Flowers

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4614

----------------------------------------------------------------------------------

较为麻烦的一道线段树题目 为了更方便处理可以把

查询左端点 查询右端点 区间查询 这三个查询分开写

最后复杂度是 $O(mlog(n))$的 $($下面的代码就是用的这个思路$)$

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cmath>
  4 #include <algorithm>
  5 using namespace std;
  6 const int N = 50010;
  7 int sum[N << 2], flag[N << 2];
  8 int t, n, m;
  9 void pushdown(int x, int tl, int tr)
 10 {
 11     if(!flag[x])
 12         return;
 13     flag[x << 1] = flag[x << 1 | 1] = flag[x];
 14     if(flag[x] & 1)
 15         sum[x << 1] = sum[x << 1 | 1] = 0;
 16     else
 17     {
 18         int mid = (tl + tr) >> 1;
 19         sum[x << 1] = mid - tl + 1;
 20         sum[x << 1 | 1] = tr - mid;
 21     }
 22     flag[x] = 0;
 23 }
 24 void update(int x, int L, int R, int tl, int tr, int y)
 25 {
 26     if(L <= tl && tr <= R)
 27     {
 28         flag[x] = y;
 29         if(y & 1)
 30             sum[x] = 0;
 31         else
 32             sum[x] = tr - tl + 1;
 33         return;
 34     }
 35     int mid = (tl + tr) >> 1;
 36     pushdown(x, tl, tr);
 37     if(L <= mid)
 38         update(x << 1, L, R, tl, mid, y);
 39     if(mid < R)
 40         update(x << 1 | 1, L, R, mid + 1, tr, y);
 41     sum[x] = sum[x << 1] + sum[x << 1 | 1];
 42 }
 43 int queryl(int x, int L, int tl, int tr)
 44 {
 45     if(!sum[x])
 46             return -1;
 47     if(tl == tr)
 48         return tl;
 49     int mid = (tl + tr) >> 1;
 50     pushdown(x, tl, tr);
 51     int re = -1;
 52     if(L <= mid && sum[x << 1])
 53         re = queryl(x << 1, L, tl, mid);
 54     if(re != -1)
 55         return re;
 56         return queryl(x << 1 | 1, L, mid + 1, tr);
 57 }
 58 int queryr(int x, int L, int tl, int tr, int &num)
 59 {
 60     if(tl == tr)
 61     {
 62         num -= sum[x];
 63         return tl;
 64     }
 65     int mid = (tl + tr) >> 1;
 66     pushdown(x, tl, tr);
 67     if(L <= tl)
 68     {
 69         if(sum[x << 1] >= num || !sum[x << 1 | 1])
 70             return queryr(x << 1, L, tl, mid, num);
 71         num -= sum[x << 1];
 72         return queryr(x << 1 | 1, L, mid + 1, tr, num);
 73     }
 74     if(L <= mid)
 75     {
 76         int re = queryr(x << 1, L, tl, mid, num);
 77         if(!num || !sum[x << 1 | 1])
 78             return re;
 79     }
 80     return queryr(x << 1 | 1, L, mid + 1, tr, num);
 81 }
 82 int query(int x, int L, int R, int tl, int tr)
 83 {
 84     if(L <= tl && tr <= R)
 85         return tr - tl + 1 - sum[x];
 86     int mid = (tl + tr) >> 1, re = 0;
 87     pushdown(x, tl, tr);
 88     if(L <= mid)
 89         re += query(x << 1, L, R, tl, mid);
 90     if(mid < R)
 91         re += query(x << 1 | 1, L, R, mid + 1, tr);
 92     return re;
 93 }
 94 int main()
 95 {
 96     scanf("%d", &t);
 97     while(t--)
 98     {
 99         scanf("%d%d", &n, &m);
100         memset(sum, 0, sizeof sum);
101         memset(flag, 0, sizeof flag);
102         update(1, 0, n - 1, 0, n - 1, 2);
103         int x, y, z;
104         while(m--)
105         {
106             scanf("%d%d%d", &x, &y, &z);
107             if(x == 1)
108             {
109                 int L = queryl(1, y, 0, n - 1);
110                 if(L == -1)
111                 {
112                     puts("Can not put any one.");
113                     continue;
114                 }    
115                 int R = queryr(1, L, 0, n - 1, z);
116                 printf("%d %d\n", L, R);
117                 update(1, L, R, 0, n - 1, 1);
118             }
119             else
120             {
121                 printf("%d\n", query(1, y, z, 0, n - 1));
122                 update(1, y, z, 0, n - 1, 2);
123             }
124         }
125         puts("");
126     }
127     return 0;
128 }

 

----------------------------------------------------------------------------------

不过如果不想写这么多判断的话也可以多写个二分 这样就只要区间查询

然而复杂度就会变成$O(mlog^2(n))$