树状数组的基本用法(板子)

写这篇博客,为了方便自己后续复制粘贴板子; 

HDU 1166

题意:单点更新,区间查询

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 #define lowbit(x) ((x)&-(x))
 7 typedef long long ll;
 8 const int maxn=5e5+5;
 9 int n, q, a[maxn], c[maxn];
10 
11 void add(int x, int val)
12 {
13     while(x<=n)
14     {
15         c[x]+=val;
16         x+=lowbit(x);
17     }
18 }
19 
20 int ask(int x)
21 {
22     int ans=0;
23     while(x>0)
24     {
25         ans+=c[x];
26         x-=lowbit(x);
27     }
28     return ans;
29 }
30 
31 int main()
32 {
33     //freopen("in.txt", "r", stdin);
34     //ios::sync_with_stdio(false);
35     int T, kase=0; cin>>T;
36     while(T--)
37     {
38         cin>>n;
39         memset(c, 0, sizeof(c));
40         for(int i=1; i<=n; i++)
41         {
42             int x; cin>>x;
43             add(i, x);
44         }
45 
46         printf("Case %d:\n", ++kase);
47         char s[10];
48         while(scanf("%s", s+1), s[1]!='E')
49         {
50             int x,y; cin>>x>>y;
51             if(s[1]=='Q')
52                 printf("%d\n", ask(y)-ask(x-1));
53             else if(s[1]=='A')
54                 add(x, y);
55             else
56                 add(x, -y);
57         }
58     }
59     return 0;
60 }
View Code

 

HDU 1754

题意:区间查询(最大值),单点更新; 

不过好像树状数组在维护最大值好像没有线段树快,这个看看就行了,遇到了还是写线段树吧!

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 #define lowbit(x) ((x)&-(x))
 7 typedef long long ll;
 8 const int maxn=5e5+5;
 9 int n, q, a[maxn], c[maxn];
10 
11 void add(int x, int val)
12 {
13     while(x<=n)
14     {
15         c[x]=val;
16         for(int i=1; i<lowbit(x); i<<=1)
17             c[x]=max(c[x], c[x-i]);
18         x+=lowbit(x);
19     }
20 }
21 
22 int ask(int l, int r)
23 {
24     int ans=0;
25     while(l<=r)
26     {
27         ans=max(ans, a[r]);   ///注意这一行是和a数组比的
28         r--;
29         for( ; r-l>=lowbit(r); r-=lowbit(r))
30             ans=max(ans, c[r]);
31     }
32     return ans;
33 }
34 
35 int main()
36 {
37     //freopen("in.txt", "r", stdin);
38     ios::sync_with_stdio(false);
39     while(cin>>n>>q)
40     {
41         for(int i=1; i<=n; i++)
42         {
43             cin>>a[i];
44             add(i, a[i]);
45         }
46 
47         while(q--)
48         {
49             char cmd; int x, y;
50             cin>>cmd>>x>>y;
51             if(cmd=='Q')
52                 printf("%d\n", ask(x,y));
53             else
54             {
55                 a[x]=y;
56                 add(x, y);
57             }
58         }
59     }
60     return 0;
61 }
View Code

 

POJ 3468 

题意:区间查询,区间更新; 

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 using namespace std;
 5 
 6 #define lowbit(x) ((x)&-(x))
 7 #define _rep(x,y) for(int i=(x); i<=(y); i++)
 8 typedef long long ll;
 9 const int maxn=1e5+5;
10 ll sum1[maxn], sum2[maxn], sum[maxn];
11 int n, q, a[maxn], c[maxn];
12 
13 void add(int x, int val)
14 {
15     int i=x;
16     while(i<=n)
17     {
18         sum1[i]+=val;
19         sum2[i]+=val*x;
20         i+=lowbit(i);
21     }
22 }
23 
24 void range_add(int l, int r, int x)
25 {
26     add(l, x); add(r+1, -x);
27 }
28 
29 ll ask(int x)
30 {
31     ll ans=0; int i=x;
32     while(i>0)
33     {
34         ans+=(x+1)*sum1[i]-sum2[i];
35         i-=lowbit(i);
36     }
37     return ans;
38 }
39 
40 ll range_ask(int l, int r)
41 {
42     return ask(r)-ask(l-1);
43 }
44 
45 
46 int main()
47 {
48     //freopen("in.txt", "r", stdin); 
49     ios::sync_with_stdio(false);
50     cin>>n>>q;
51     _rep(1,n){
52         cin>>a[i];
53         sum[i]=sum[i-1]+a[i];
54     }
55     while(q--)
56     {
57         char cmd; int x,y;
58         cin>>cmd>>x>>y;
59         if(cmd=='Q')
60             printf("%I64d\n", sum[y]-sum[x-1]+range_ask(x,y));
61         else
62         {
63             int z; cin>>z;
64             range_add(x, y, z);
65         }
66     }
67     return 0;
68 }
View Code

 

HDU 1698 (未解决)

题意:区间覆盖(不会),区间查询

 

posted @ 2019-05-03 12:09  N_Yokel  阅读(258)  评论(0编辑  收藏  举报