分块来水题

luogu P3374 【模板】树状数组 1

 

在大牛分站交能过，主站卡常。

时间复杂度为 n√n ≈ 3.5 * 108，我都不知道怎么过的。。

 

——代码

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <iostream>
 4 
 5 using namespace std;
 6 
 7 const int MAXN = 500001;
 8 int n, m, S, C;
 9 int a[MAXN], st[MAXN], ed[MAXN], belong[MAXN], sum[MAXN];
10 
11 inline void init()
12 {
13     int i, j;
14     scanf("%d %d", &n, &m);
15     for(i = 1; i <= n; i++) scanf("%d", &a[i]);
16     S = sqrt(n) + 10;
17     for(i = 1; i <= n; i += S)
18     {
19         st[++C] = i;
20         ed[C] = min(i + S - 1, n);
21     }
22     for(i = 1; i <= C; i++)
23         for(j = st[i]; j <= ed[i]; j++)
24             sum[i] += a[j], belong[j] = i;
25 }
26 
27 inline void update(int x, int k)
28 {
29     a[x] += k;
30     sum[belong[x]] += k;
31 }
32 
33 inline int query(int x, int y)
34 {
35     int i, ans = 0, l = belong[x], r = belong[y];
36     if(l == r)
37     {
38         for(i = x; i <= y; i++) ans += a[i];
39         return ans;
40     }
41     for(i = x; i <= ed[l]; i++) ans += a[i];
42     for(i = l + 1; i <= r - 1; i++) ans += sum[i];
43     for(i = st[r]; i <= y; i++) ans += a[i];
44     return ans;
45 }
46 
47 inline void work()
48 {
49     int i, x, y, z;
50     for(i = 1; i <= m; i++)
51     {
52         scanf("%d %d %d", &z, &x, &y);
53         if(z == 1) update(x, y);
54         else printf("%d\n", query(x, y));
55     }
56 }
57 
58 int main()
59 {
60     init();
61     work();
62     return 0;
63 }
View Code

 

 

luogu P3368 【模板】树状数组 2

 

在大牛分站交能过，主站卡常。

 

——代码

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <iostream>
 4 
 5 using namespace std;
 6 
 7 const int MAXN = 500001;
 8 int n, m, S, C;
 9 int a[MAXN], belong[MAXN], add[MAXN], st[MAXN], ed[MAXN];
10 
11 inline void init()
12 {
13     int i, j;
14     scanf("%d %d", &n, &m);
15     for(i = 1; i <= n; i++) scanf("%d", &a[i]);
16     S = sqrt(n);
17     for(i = 1; i <= n; i += S)
18     {
19         st[++C] = i;
20         ed[C] = min(i + S - 1, n);
21     }
22     for(i = 1; i <= C; i++)
23         for(j = st[i]; j <= ed[i]; j++)
24             belong[j] = i;
25 }
26 
27 inline void update(int x, int y, int k)
28 {
29     int i, l = belong[x], r = belong[y];
30     if(l == r)
31     {
32         for(i = x; i <= y; i++) a[i] += k;
33         return;
34     }
35     for(i = x; i <= ed[l]; i++) a[i] += k;
36     for(i = l + 1; i <= r - 1; i++) add[i] += k;
37     for(i = st[r]; i <= y; i++) a[i] += k;
38 }
39 
40 inline void work()
41 {
42     int i, x, y, z, d;
43     for(i = 1; i <= m; i++)
44     {
45         scanf("%d %d", &z, &x);
46         if(z == 1)
47         {
48             scanf("%d %d", &y, &d);
49             update(x, y, d);
50         }
51         else printf("%d\n", a[x] + add[belong[x]]);
52     }
53 }
54 
55 int main()
56 {
57     init();
58     work();
59     return 0;
60 }
View Code

 

 

luogu P3372 【模板】线段树 1

 

这个数据水，主站就能过。

 

——代码

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <iostream>
 4 #define LL long long
 5 
 6 using namespace std;
 7 
 8 const int MAXN = 100001;
 9 int n, m, S, C;
10 int st[MAXN], ed[MAXN], belong[MAXN];
11 LL a[MAXN], sum[MAXN], add[MAXN];
12 
13 inline void init()
14 {
15     int i, j;
16     scanf("%d %d", &n, &m);
17     for(i = 1; i <= n; i++) scanf("%lld", &a[i]);
18     S = sqrt(n);
19     for(i = 1; i <= n; i += S)
20     {
21         st[++C] = i;
22         ed[C] = min(i + S - 1, n);
23     }
24     for(i = 1; i <= C; i++)
25         for(j = st[i]; j <= ed[i]; j++)
26             sum[i] += a[j], belong[j] = i;
27 }
28 
29 inline void update(int x, int y, LL k)
30 {
31     int i, l = belong[x], r = belong[y];
32     if(l == r)
33     {
34         for(i = x; i <= y; i++) a[i] += k;
35         sum[l] += (y - x + 1) * k;
36         return;
37     }
38     for(i = x; i <= ed[l]; i++) a[i] += k;
39     sum[l] += (ed[l] - x + 1) * k;
40     for(i = l + 1; i <= r - 1; i++)
41     {
42         sum[i] += (ed[i] - st[i] + 1) * k;
43         add[i] += k;
44     }
45     for(i = st[r]; i <= y; i++) a[i] += k;
46     sum[r] += (y - st[r] + 1) * k;
47 }
48 
49 inline LL query(int x, int y)
50 {
51     int i, l = belong[x], r = belong[y];
52     LL ans = 0;
53     if(l == r)
54     {
55         for(i = x; i <= y; i++) ans += a[i] + add[l];
56         return ans;
57     }
58     for(i = x; i <= ed[l]; i++) ans += a[i] + add[l];
59     for(i = l + 1; i <= r - 1; i++) ans += sum[i];
60     for(i = st[r]; i <= y; i++) ans += a[i] + add[r];
61     return ans;
62 }
63 
64 inline void work()
65 {
66     int i, x, y, z;
67     LL k;
68     for(i = 1; i <= m; i++)
69     {
70         scanf("%d %d %d", &z, &x, &y);
71         if(z == 1)
72         {
73             scanf("%lld", &k);
74             update(x, y, k);
75         }
76         else printf("%lld\n", query(x, y));
77     }
78 }
79 
80 int main()
81 {
82     init();
83     work();
84     return 0;
85 }
View Code

 

luogu P3373 【模板】线段树 2

 

看了黄学长的讲解，猛然醒悟！%%%%%hzwer

整块可以直接打标记，而左右两边零散的块要先把它们所属于的整块的标记清除掉再打，否则会错。

 

——代码

  1 #include <cmath>
  2 #include <cstdio>
  3 #include <iostream>
  4 #define LL long long
  5 
  6 using namespace std;
  7 
  8 const int MAXN = 100001;
  9 int n, m, p, S, C;
 10 int st[MAXN], ed[MAXN], belong[MAXN];
 11 LL a[MAXN], sum[MAXN], add[MAXN], mul[MAXN];
 12 
 13 inline void init()
 14 {
 15     int i, j;
 16     scanf("%d %d %d", &n, &m, &p);
 17     for(i = 1; i <= n; i++) scanf("%lld", &a[i]);
 18     S = sqrt(n);
 19     for(i = 1; i <= n; i += S)
 20     {
 21         st[++C] = i;
 22         ed[C] = min(i + S - 1, n);
 23         mul[C] = 1;
 24     }
 25     for(i = 1; i <= C; i++)
 26         for(j = st[i]; j <= ed[i]; j++)
 27             sum[i] += a[j], belong[j] = i;
 28 }
 29 
 30 inline void reset(int x)
 31 {
 32     int i;
 33     for(i = st[x]; i <= ed[x]; i++)
 34         a[i] = (a[i] * mul[x] + add[x]) % p;
 35     sum[x] = (sum[x] * mul[x] + add[x] * (ed[x] - st[x] + 1)) % p;
 36     add[x] = 0, mul[x] = 1;
 37 }
 38 
 39 inline void update(int f, int x, int y, LL k)
 40 {
 41     int i, l = belong[x], r = belong[y];
 42     LL tot;
 43     if(l == r)
 44     {
 45         reset(l);
 46         for(tot = 0, i = x; i <= y; i++)
 47         {
 48             if(f == 1) a[i] = (a[i] * k) % p, tot = (tot + a[i]) % p;
 49             else a[i] = (a[i] + k) % p;
 50         }
 51         if(f == 1) sum[l] = (sum[l] + tot * (k - 1)) % p;
 52         else sum[l] = (sum[l] + (y - x + 1) * k) % p;
 53         return;
 54     }
 55     
 56     reset(l);
 57     for(tot = 0, i = x; i <= ed[l]; i++)
 58     {
 59         if(f == 1) tot = (tot + a[i]) % p, a[i] = (a[i] * k) % p;
 60         else a[i] = (a[i] + k) % p;
 61     }
 62     if(f == 1) sum[l] = (sum[l] + tot * (k - 1)) % p;
 63     else sum[l] = (sum[l] + (ed[l] - x + 1) * k) % p;
 64     
 65     reset(r);
 66     for(tot = 0, i = st[r]; i <= y; i++)
 67     {
 68         if(f == 1) tot = (tot + a[i]) % p, a[i] = (a[i] * k) % p;
 69         else a[i] = (a[i] + k) % p;
 70     }
 71     if(f == 1) sum[r] = (sum[r] + tot * (k - 1)) % p;
 72     else sum[r] = (sum[r] + (y - st[r] + 1) * k) % p;
 73     
 74     for(i = l + 1; i <= r - 1; i++)
 75         if(f == 1)
 76         {
 77             add[i] = (add[i] * k) % p;
 78             mul[i] = (mul[i] * k) % p;
 79         }
 80         else add[i] = (add[i] + k) % p;
 81 }
 82 
 83 inline LL query(int x, int y)
 84 {
 85     int i, l = belong[x], r = belong[y];
 86     LL ans = 0;
 87     if(l == r)
 88     {
 89         for(i = x; i <= y; i++)
 90             ans = (ans + a[i] * mul[l] + add[l]) % p;
 91         return ans;
 92     }
 93     for(i = x; i <= ed[l]; i++)
 94         ans = (ans + a[i] * mul[l] + add[l]) % p;
 95     for(i = l + 1; i <= r - 1; i++)
 96         ans = (ans + sum[i] * mul[i] + add[i] * (ed[i] - st[i] + 1)) % p;
 97     for(i = st[r]; i <= y; i++)
 98         ans = (ans + a[i] * mul[r] + add[r]) % p;
 99     return ans;
100 }
101 
102 inline void work()
103 {
104     int i, x, y, z;
105     LL k;
106     for(i = 1; i <= m; i++)
107     {
108         scanf("%d %d %d", &z, &x, &y);
109         if(z == 3) printf("%lld\n", query(x, y));
110         else scanf("%lld", &k), update(z, x, y, k);
111     }
112 }
113 
114 int main()
115 {
116     init();
117     work();
118     return 0;
119 }
View Code

 

 

 

还有一个链接，黄学长的分块姿势。

【分块】数列分块入门1-9 by hzwer

posted @ 2017-05-08 10:49  zht467  阅读(142)  评论(0)    收藏  举报