A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 73163 | Accepted: 22585 | |
| 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
Hint
The sums may exceed the range of 32-bit integers.
Source
POJ Monthly--2007.11.25, Yang Yi
最裸的线段树lazy标记
1 #include <iostream>
2 #include <cstdio>
3 using namespace std;
4 const int N = 100005;
5 typedef long long LL;
6 LL sum[N<<2]; //sum用来存储每个节点的子节点数值的总和
7 LL add[N<<2];//add用来记录该节点的每个数值应该加多少
8 struct Node{
9 int l,r;//表示改点的左右区间
10 int mid(){//结构体函数
11 return (l+r)>>1;
12 }
13 } tree[N<<2];
14
15 void PushUp(int rt){//算某一节点的左右孩子值的和
16 sum[rt] = sum[rt<<1] + sum[rt<<1|1];
17 }
18
19 void PushDown(int rt,int m){//rt当前节点 m此节点的区间长度
20 if(add[rt]!=0){//如果当前节点lazy标记不为 0
21 add[rt<<1] += add[rt];//左右孩子lazy累加
22 add[rt<<1|1] += add[rt];
23 sum[rt<<1] += add[rt]*(m-(m>>1));//更新计算左右孩子
24 sum[rt<<1|1] += add[rt] * (m>>1);
25 add[rt] = 0;//小细节，但很重要，不能忘记lazy用过后清零
26 }
27 }
28
29 void build(int l,int r,int rt){
30 tree[rt].l = l;
31 tree[rt].r = r;
32 add[rt] = 0;//lazy标记记为0
33 if(l == r){
34 scanf("%lld",&sum[rt]);//把子节点信息记录在sum里
35 return ;
36 }
37 int m = tree[rt].mid();
38 build(l,m,rt<<1);//建立左右孩子，这时编号是按照类似“宽搜”来编的
39 build(m+1,r,rt<<1|1);
40
41 PushUp(rt);//算出此节点的权值
42 }
43
44 void update(int c,int l,int r,int rt){//当前节点rt，在l和r区间上加上c
45 if(l<=tree[rt].l&&tree[rt].r<=r){//当前节点所表示的区间完全被所要更新的区间包含
46 add[rt]+=c;//lazy累加,add[i]数组是针对i左右孩子的
47 sum[rt]+=(LL)c*(tree[rt].r-tree[rt].l+1);//这句话很重要和下面的PushUP()类似
48 return;
49 }
50 PushDown(rt,tree[rt].r - tree[rt].l + 1);//用rt的lazy更新其子节点
51 int m = tree[rt].mid();
52 if(l<=m) update(c,l,r,rt<<1);
53 if(m+1<=r) update(c,l,r,rt<<1|1);
54 PushUp(rt);
55 }
56
57 LL query(int l,int r,int rt){//当前节点为rt，求l，到r的和
58 if(l<=tree[rt].l&&tree[rt].r<=r){//区间完全包含，可以直接返回
59 return sum[rt];
60 }
61 //因为此时用到rt节点，所以才更新lazy
62 PushDown(rt,tree[rt].r-tree[rt].l + 1);
63
64 int m = tree[rt].mid();
65 LL res = 0;
66
67 if(l<= m)//要求的区间有一部分在mid的左边
68 res += query(l,r,rt<<1);
69 if(m+1<=r)
70 res += query(l,r,rt<<1|1);
71 return res;
72 }
73
74 int main(){
75
76 int n,m;
77
78 scanf("%d %d",&n,&m);
79 build(1,n,1);
80
81 while(m--){
82 char ch[2];
83 scanf("%s",ch);
84 int a,b,c;
85 if(ch[0] == 'Q'){
86 scanf("%d %d", &a,&b);
87 printf("%lld
",query(a,b,1));
88 }
89 else{
90 scanf("%d %d %d",&a,&b,&c);
91 cin>>a>>b>>c;
92 update(c,a,b,1);
93 }
94 }
95
96 return 0;
97 }
还有一种较好理解的方法：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cmath> 5 #include<algorithm> 6 #include<cstring> 7 using namespace std; 8 typedef long long LL; 9 inline LL read(){ 10 LL x=0,f=1;char ch=getchar(); 11 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 12 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 13 return x*f; 14 } 15 const LL maxn=100005; 16 struct node{ 17 LL num; 18 LL l,r; 19 LL lc,rc; 20 LL sum,lazy; 21 }seg[maxn*8]; 22 LL a[maxn]; 23 LL N,M,tot=1; 24 inline void updata_son(LL root){ 25 LL d=seg[root].lazy; 26 if(d!=0){ 27 LL lson=seg[root].lc; LL rson=seg[root].rc; 28 seg[lson].lazy+=d; seg[rson].lazy+=d; seg[root].lazy=0; 29 seg[lson].sum+=d*(seg[lson].r-seg[lson].l+1); 30 seg[rson].sum+=d*(seg[rson].r-seg[rson].l+1); 31 32 } 33 } 34 inline void Build(LL root,LL l,LL r){ 35 seg[root].num=root; 36 seg[root].l=l; seg[root].r=r; 37 if(l==r){ 38 seg[root].sum=a[l]; 39 return ; 40 } 41 LL mid=(l+r)>>1; 42 seg[root].lc=++tot; Build(tot,l,mid); 43 seg[root].rc=++tot; Build(tot,mid+1,r); 44 seg[root].sum=seg[seg[root].lc].sum+seg[seg[root].rc].sum; 45 } 46 inline LL query(LL root,LL l,LL r){ 47 if(l<=seg[root].l&&seg[root].r<=r){ 48 return seg[root].sum; 49 } 50 updata_son(root); 51 LL ans=0; 52 LL mid=(seg[root].l+seg[root].r)>>1; 53 if(l<=mid) ans+=query(seg[root].lc,l,r); 54 if(mid+1<=r) ans+=query(seg[root].rc,l,r); 55 return ans; 56 } 57 58 inline void change(LL root,LL l,LL r,LL delta){ 59 if(l<=seg[root].l&&seg[root].r<=r){ 60 seg[root].sum+=(seg[root].r-seg[root].l+1)*delta; 61 seg[root].lazy+=delta; 62 return ; 63 } 64 updata_son(root); 65 LL mid=(seg[root].l+seg[root].r)>>1; 66 if(l<=mid) change(seg[root].lc,l,r,delta); 67 if(mid+1<=r) change(seg[root].rc,l,r,delta); 68 seg[root].sum=seg[seg[root].lc].sum+seg[seg[root].rc].sum; 69 } 70 int main(){ 71 N=read(); M=read(); 72 for(LL i=1;i<=N;i++) a[i]=read(); 73 Build(1,1,N); 74 LL ll,rr,de; 75 char x[2]; 76 for(LL i=1;i<=M;i++){ 77 scanf("%s",x); 78 if(x[0]=='C'){ 79 ll=read(); rr=read(); de=read(); 80 change(1,ll,rr,de); 81 continue; 82 } 83 if(x[0]=='Q'){ 84 ll=read(); rr=read(); 85 printf("%lld ",query(1,ll,rr)); 86 } 87 } 88 return 0; 89 }