kuangbin专题七 POJ3468 A Simple Problem with Integers （线段树或树状数组）
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
区间修改 区间查询 注意pushdown的操作就可以了
1 #include <iostream> 2 #include <stdio.h> 3 #include <math.h> 4 #include <string.h> 5 #include <stdlib.h> 6 #include <string> 7 #include <vector> 8 #include <set> 9 #include <map> 10 #include <queue> 11 #include <algorithm> 12 #include <sstream> 13 #include <stack> 14 using namespace std; 15 #define FO freopen("in.txt","r",stdin); 16 #define rep(i,a,n) for (int i=a;i<n;i++) 17 #define per(i,a,n) for (int i=n-1;i>=a;i--) 18 #define pb push_back 19 #define mp make_pair 20 #define all(x) (x).begin(),(x).end() 21 #define fi first 22 #define se second 23 #define SZ(x) ((int)(x).size()) 24 #define debug(x) cout << "&&" << x << "&&" << endl; 25 #define lowbit(x) (x&-x) 26 #define mem(a,b) memset(a, b, sizeof(a)); 27 typedef vector<int> VI; 28 typedef long long ll; 29 typedef pair<int,int> PII; 30 const ll mod=1000000007; 31 const int inf = 0x3f3f3f3f; 32 ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} 33 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 34 //head 35 36 //区间修改 区间查询 37 const int maxx=100010; 38 ll sum[maxx<<2],lazy[maxx<<2],val; 39 int n,q; 40 41 void Pushup(int rt) { 42 sum[rt]=sum[rt<<1]+sum[rt<<1|1]; 43 } 44 45 void build(int rt,int L,int R) { 46 lazy[rt]=0; 47 if(L==R) { 48 scanf("%lld",&sum[rt]); 49 return; 50 } 51 int mid=(L+R)>>1; 52 build(rt<<1,L,mid); 53 build(rt<<1|1,mid+1,R); 54 Pushup(rt); 55 } 56 57 void Pushdown(int rt,int x) { 58 if(lazy[rt]) { 59 lazy[rt<<1]+=lazy[rt]; 60 lazy[rt<<1|1]+=lazy[rt]; 61 sum[rt<<1]+=(x-(x>>1))*lazy[rt];//左子树加左边一半的值 62 sum[rt<<1|1]+=(x>>1)*lazy[rt];//右子树同理 63 lazy[rt]=0;//清空 64 } 65 } 66 67 void Updata(int rt,int L,int R,int l,int r) { 68 if(L>=l&&R<=r) { 69 lazy[rt]+=val;//累加标记 70 sum[rt]+=(R-L+1)*val;//更新值 71 return; 72 } 73 int mid=(L+R)>>1; 74 Pushdown(rt,R-L+1);//这里多了一个参数 L R区间的个数 75 if(l<=mid) Updata(rt<<1,L,mid,l,r); 76 if(r>mid) Updata(rt<<1|1,mid+1,R,l,r); 77 Pushup(rt); 78 } 79 80 ll Query(int rt,int L,int R,int l,int r) { 81 if(L>=l&&R<=r) 82 return sum[rt]; 83 ll ans=0; 84 int mid=(L+R)>>1; 85 Pushdown(rt,R-L+1); 86 if(l<=mid) ans+=Query(rt<<1,L,mid,l,r); 87 if(r>mid) ans+=Query(rt<<1|1,mid+1,R,l,r); 88 Pushup(rt); 89 return ans; 90 } 91 92 93 int main() { 94 while(~scanf("%d%d",&n,&q)) { 95 build(1,1,n); 96 char s[3]; 97 while(q--) { 98 scanf("%s",s); 99 int l,r; 100 if(s[0]=='Q') { 101 scanf("%d%d",&l,&r); 102 printf("%lld\n",Query(1,1,n,l,r)); 103 } else { 104 scanf("%d%d%lld",&l,&r,&val); 105 Updata(1,1,n,l,r); 106 } 107 } 108 } 109 }
借此机会学习了一波树状数组，有点难理解。
单点修改 区间查询 Updata(l,val)
区间修改 单点查询 引入差分数组 c[i]=d[i]-d[i-1]; Updata(l,val); Updata(r+1,-val); ----明白树状数组就好理解了。把多加的减去
区间修改 区间查询 公式可以推到一下，sum[n]=n*(c[1]+c[2]+...c[n])-(0*c[1]+1*c[2]+2*c[3]+...+(n-1)*c[n]);
所以再维护一个 (i-1)*(a[i]-a[i-1]) , a为原数组。
Updata(l,(l-1)*val); Updata(r+1,-r*val); 原理同上。多加的减去。
（纯属个人理解。关键还是公式的推导，和树状数组的理解）
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 typedef long long ll; 6 #define lowbit(x) (x&-x) 7 8 int n,q; 9 const int maxn=100010; 10 ll a[maxn],val,c1[maxn],c2[maxn];//维护两个树状数组 11 12 void Updata(ll c[],int pos,ll val) { 13 while(pos<=n) { 14 c[pos]+=val; 15 pos+=lowbit(pos); 16 } 17 } 18 19 ll getsum(ll c[],int pos) { 20 ll ans=0; 21 while(pos>0) { 22 ans+=c[pos]; 23 pos-=lowbit(pos); 24 } 25 return ans; 26 } 27 28 ll sigma(int r) {//公式推导 29 ll sum1=r*getsum(c1,r); 30 ll sum2=getsum(c2,r); 31 return sum1-sum2; 32 } 33 34 ll Query(int l,int r) {//区间求和 35 return sigma(r)-sigma(l-1); 36 } 37 38 int main() { 39 while(~scanf("%d%d",&n,&q)) { 40 memset(c1,0,sizeof(c1));memset(c2,0,sizeof(c2)); 41 for(int i=1;i<=n;i++) { 42 scanf("%lld",&a[i]); 43 Updata(c1,i,a[i]-a[i-1]);//维护a[i]-a[i-1] 44 Updata(c2,i,(i-1)*(a[i]-a[i-1]));//维护(i-1)*(a[i]-a[i-1]) 45 } 46 char s[2]; 47 while(q--) { 48 int l,r; 49 scanf("%s",s); 50 if(s[0]=='Q') { 51 scanf("%d%d",&l,&r); 52 printf("%lld\n",Query(l,r)); 53 } else { 54 scanf("%d%d%lld",&l,&r,&val); 55 //更新操作有点难理解 56 Updata(c1,l,val);Updata(c1,r+1,-val); 57 Updata(c2,l,(l-1)*val);Updata(c2,r+1,-r*val); 58 } 59 } 60 } 61 }
