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
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）.
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 109930 | Accepted: 34223 | |
| Case Time Limit: 2000MS | ||
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
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
#include
#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
当N = 0，输入结束。
#include
#include
#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;
}