POJ 3468 线段树 成段更新 懒惰标记
A Simple Problem with Integers
Time Limit:5000MS Memory Limit:131072K
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.
题解：本题也是线段树系列的模板题之一，要求的是成段更新+懒惰标记。PS：原题的说明有问题，上面说的“C a b c”中的c的范围其实在int32之外，需要使用long long，否则定是WA，真心坑爹，连WA多次，还是在discuss中看到的原因。
稍微讲解下代码中的一些细节: step<<1 与 step<<1|1，意思分别是step*2 和step*+1，具体为什么，可以回去复习一下位运算
1 #include<iostream> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cstdio> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<map> 9 10 #define N 100010 11 #define M 15 12 #define mod 1000000007 13 #define mod2 100000000 14 #define ll long long 15 #define maxi(a,b) (a)>(b)? (a) : (b) 16 #define mini(a,b) (a)<(b)? (a) : (b) 17 18 using namespace std; 19 20 int n,q; 21 int x[N]; 22 char ch; 23 int a,b,c; 24 25 struct node 26 { 27 ll mark,sum; 28 }tree[4*N]; 29 30 void update(int l,int r,int i) 31 { 32 if(!tree[i].mark) return; 33 int mid=(l+r)/2; 34 tree[2*i].sum+=tree[i].mark*(ll)(mid-l+1); 35 tree[2*i+1].sum+=tree[i].mark*(ll)(r-mid); 36 tree[2*i].mark+=tree[i].mark; 37 tree[2*i+1].mark+=tree[i].mark; 38 tree[i].mark=0; 39 40 } 41 42 ll query(int tl,int tr,int l,int r,int i) 43 { 44 if(tl>r || tr<l) 45 return 0; 46 if(tl<=l && r<=tr) 47 return tree[i].sum; 48 update(l,r,i); 49 int mid=(l+r)/2; 50 51 return query(tl,tr,l,mid,2*i)+query(tl,tr,mid+1,r,2*i+1); 52 } 53 54 void addd(int tl,int tr,int l,int r,int i,int val) 55 { 56 if(tl>r || tr<l) 57 return; 58 if(tl<=l && tr>=r){ 59 tree[i].sum+=val*(ll)(r-l+1); 60 tree[i].mark+=val; 61 return; 62 } 63 update(l,r,i); 64 int mid=(l+r)/2; 65 addd(tl,tr,l,mid,2*i,val); 66 addd(tl,tr,mid+1,r,2*i+1,val); 67 tree[i].sum=tree[2*i].sum+tree[2*i+1].sum; 68 } 69 70 void build_tree(int l,int r,int i) 71 { 72 if(l==r){ 73 tree[i].sum=x[l]; 74 return; 75 } 76 int mid=(l+r)/2; 77 build_tree(l,mid,2*i); 78 build_tree(mid+1,r,2*i+1); 79 tree[i].sum=tree[2*i].sum+tree[2*i+1].sum; 80 } 81 82 int main() 83 { 84 int i; 85 // freopen("data.in","r",stdin); 86 //scanf("%d",&T); 87 //for(int cnt=1;cnt<=T;cnt++) 88 //while(T--) 89 while(scanf("%d%d",&n,&q)!=EOF) 90 { 91 // printf(" %d %d \n",n,q); 92 for(i=1;i<=n;i++) scanf("%d",&x[i]); 93 // printf(" 1\n"); 94 memset(tree,0,sizeof(tree)); 95 build_tree(1,n,1); 96 97 98 // printf(" 2\n"); 99 getchar(); 100 while(q--){ 101 // printf(" 3\n"); 102 scanf("%c",&ch); 103 // printf(" %c\n",ch); 104 if(ch=='C'){ 105 //printf(" 31\n"); 106 scanf("%d%d%d",&a,&b,&c); 107 getchar(); 108 addd(a,b,1,n,1,c); 109 110 } 111 else{ 112 // printf(" 32\n"); 113 scanf("%d%d",&a,&b); 114 getchar(); 115 ll ans=query(a,b,1,n,1); 116 printf("%I64d\n",ans); 117 } 118 } 119 } 120 121 return 0; 122 }
