解释下关于数状数组区间更新、单点查询和区间更新、区间查询
2017-06-14 21:03
316 查看
首先说明一点,HDU - 1556 Color the ball (一维树状数组 + 区间修改 + 单点求值),比如给区间[a, b]加1,就直接用,add(a, 1),add(b+1, 1),那是因为这个题本来就满足前缀和,可以这么用,对于其他的序列这样用肯定是错的。现在来讲一下树状数组正确的区间更新,好坑啊!网上找了很多博客和文章,讲的都不清楚,有的还以为像气球涂色一样,这样做就是正确的区间更新。所以决定写篇文章好好讲一下树状数组的区间更新。
对原数组做拆分,即令的d[i] = a[i] – a[i-1],特别地d[i] = a[i].则a[i] = d[1] + d[2] + … + d[i];
做单点查询就是在求d[1….i]的和。给整个区间[l, r]增加k,d[l] += k, d[r+1] -= k;维护的d数组就行了。区间求和,仍然沿用d数组,考虑a数组[1,x]区间和的计算。d[1]被累加了x次,d[2]被累加了x-1次,...,d[x]被累加了1次。sigma(a[i]) =sigma{d[i]*(x-i+1)}
=sigma{ d[i]*(x+1) - d[i]*i } =(x+1)*sigma(d[i])-sigma(d[i]*i)。相当于把每个的d[i]都加x+1次,然后减去多加的。如果区间[r,x],用两个和减一下。sum=(x+1)*sigma(d[i])-sigma(d[i]*i)-(r)*sigma(d[i])+sigma(d[i]*i)。[请注意我们上面的讨论都假定了a[]初始全是0。如果不是这样呢?下面的程序里给出了一个相对简便的处理办法。]我们来手撕一道例题POJ3468(区间更新,区间查询).
A Simple Problem with Integers
Description
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is
to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
Sample Output
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20225 Accepted Submission(s): 10074
Problem Description
N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了,你能帮他算出每个气球被涂过几次颜色吗?
Input
每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行,每行包括2个整数a b(1 <= a <= b <= N)。
当N = 0,输入结束。
Output
每个测试实例输出一行,包括N个整数,第I个数代表第I个气球总共被涂色的次数。
Sample Input
3
1 1
2 2
3 3
3
1 1
1 2
1 3
0
Sample Output
1 1 13 2 1
对原数组做拆分,即令的d[i] = a[i] – a[i-1],特别地d[i] = a[i].则a[i] = d[1] + d[2] + … + d[i];
做单点查询就是在求d[1….i]的和。给整个区间[l, r]增加k,d[l] += k, d[r+1] -= k;维护的d数组就行了。区间求和,仍然沿用d数组,考虑a数组[1,x]区间和的计算。d[1]被累加了x次,d[2]被累加了x-1次,...,d[x]被累加了1次。sigma(a[i]) =sigma{d[i]*(x-i+1)}
=sigma{ d[i]*(x+1) - d[i]*i } =(x+1)*sigma(d[i])-sigma(d[i]*i)。相当于把每个的d[i]都加x+1次,然后减去多加的。如果区间[r,x],用两个和减一下。sum=(x+1)*sigma(d[i])-sigma(d[i]*i)-(r)*sigma(d[i])+sigma(d[i]*i)。[请注意我们上面的讨论都假定了a[]初始全是0。如果不是这样呢?下面的程序里给出了一个相对简便的处理办法。]我们来手撕一道例题POJ3468(区间更新,区间查询).
A Simple Problem with Integers
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 109930 | Accepted: 34223 | |
Case Time Limit: 2000MS |
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is
to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4
Sample Output
4 55 9 15
#include <stdio.h> #include <string.h> #define maxn 100010 long long a[maxn],b[maxn],c[maxn]; //b数组维护前缀和,c数组维护区间和 int n,m; int lowbit(const int &x){ return x&(-x); } long long query(long long *a, int x){ long long sum=0; while (x) { sum += a[x]; x -= lowbit(x); } return sum; } void update(long long *a, int x, long long w){ while ( x<= n) { a[x] += w; x += lowbit(x); } } int main(){ int l,r,i; long long ans,w; char ch; scanf("%d%d",&n,&m); a[0]=0; for(i = 1;i <= n; ++i){ scanf("%I64d",&a[i]); a[i] += a[i-1]; } while(m--){ scanf("%c",&ch); while(ch != 'Q' && ch != 'C') scanf("%c",&ch); if(ch == 'Q'){ scanf("%d%d",&l,&r); ans = a[r] - a[l - 1] + (r+1) * query(b,r) - query(c,r) - l * query(b, l-1) + query(c, l - 1); //利用起点和终点差,求区间和 printf("%I64d\n",ans); }else{ scanf("%d%d%I64d", &l, &r, &w); update(b, l, w); update(b, r + 1, -w); //前缀和更新 update(c, l, w * l); update(c, r + 1, -(r + 1) * w); //区间和更新 } } return 0; }
Color the ball
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 20225 Accepted Submission(s): 10074
Problem Description
N个气球排成一排,从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了,你能帮他算出每个气球被涂过几次颜色吗?
Input
每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行,每行包括2个整数a b(1 <= a <= b <= N)。
当N = 0,输入结束。
Output
每个测试实例输出一行,包括N个整数,第I个数代表第I个气球总共被涂色的次数。
Sample Input
3
1 1
2 2
3 3
3
1 1
1 2
1 3
0
Sample Output
1 1 13 2 1
#include <stdio.h> #include <string.h> #define maxn 100010 long long a[maxn],b[maxn],c[maxn]; //b数组维护前缀和,c数组维护区间和 int n; int lowbit(const int &x){ return x&(-x); } long long query(long long *a, int x){ long long sum=0; while (x) { sum += a[x]; x -= lowbit(x); } return sum; } void update(long long *a, int x, long long w){ while ( x<= n) { a[x] += w; x += lowbit(x); } } int main() { int l, r, ans; memset(a, 0, sizeof(a)); memset(c, 0, sizeof(c)); while(scanf("%d", &n) && n) { memset(b, 0, sizeof(b)); for (int i = 0; i < n; i++) { scanf("%d%d",&l,&r); update(b, l, 1); update(b, r + 1, -1); //前缀和更新 } for (int i = 1; i <= n; i++) { if (i == n) { printf("%I64d\n",query(b, i)); } else { printf("%I64d ",query(b, i)); } } } return 0; }用上题代码直接修改了下,就AC了。用为气球涂色是区间更新+单点查询。
相关文章推荐
- Matrix(二维数状数组+求和+区间更新+单点查询)
- 树状数组单点更新和区间查询
- hdu2642二维树状数组单点更新+区间查询
- hdu4267A Simple Problem with Integers【树状数组区间更新/单点查询】
- (树状数组)hdu1556 Color the ball(区间更新,单点查询)
- poj 2155 二维树状数组/区间更新单点查询
- hdu 4533(树状数组区间更新+单点查询)
- 树状数组基本模版(区间更新,单点查询)
- 树状数组单点更新和区间更新,二维数组poj2155(区间更新,单点查询)(已加入区间修改区间查询)
- POJ 1195 - Mobile phones 二维树状数组(单点更新..区间查询)
- 树状数组:HDU1166敌兵布阵 【单点更新,区间查询】
- HDU 1556 Color the ball (一维树状数组,区间更新,单点查询)
- HDu 1556 Color the ball【线段树&&树状数组】区间更新,单点查询
- hdu 2642 二维树状数组 单点更新区间查询 模板水题
- HDU1754线段树单点更新区间查询(数组版)
- 树状数组:HDU1556 Color the ball 【区间更新,单点查询]
- [HDU 4031]Attack[树状数组区间更新单点查询]
- POJ 2155 - Matrix 二维树状数组..区间更新..单点查询
- poj2155-二维树状数组 区间更新 单点查询
- hdu 2642 二维树状数组 单点更新区间查询 模板水题