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线段树简单区间更新模板+练习

POJ-3468 - A Simple Problem with Integers（线段树区间更新模板）

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

翻译：

给定长度为N的数列A，然后输入M行操作指令。

第一类指令形如“C a b c”，表示把数列中第a~b个数都加c。

第二类指令形如“Q a b”，表示询问数列中第a~b个数的值的和。

对于每个询问，输出一个整数表示答案。

输入格式

第一行包含两个整数N和M。

第二行包含N个整数A[i]。

接下来M行表示M条指令，每条指令的格式如题目描述所示。

输出格式

对于每个询问，输出一个整数表示答案。

每个答案占一行。

数据范围

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

Hint

The sums may exceed the range of 32-bit integers.

线段树区间更新求和的模板题，直接上代码

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <iostream>
  4 #include <string>
  5 #include <math.h>
  6 #include <algorithm>
  7 #include <vector>
  8 #include <queue>
  9 #include <set>
 10 #include <map>
 11 #include <math.h>
 12 const int INF=0x3f3f3f3f;
 13 typedef long long LL;
 14 const int mod=1e9+7;
 15 //const double PI=acos(-1);
 16 const int maxn=1e5+10;
 17 using namespace std;
 18 //ios::sync_with_stdio(false);
 19 //    cin.tie(NULL);
 20 
 21 int n,m;
 22 struct node
 23 {
 24     int l;
 25     int r;
 26     LL lazy;//注意lazy也要开LL 
 27     LL sum;
 28 }SegTree[maxn<<2];
 29 
 30 void PushUp(int rt)
 31 {
 32     SegTree[rt].sum=SegTree[rt<<1].sum+SegTree[rt<<1|1].sum;
 33 }
 34 
 35 void PushDown(int rt)
 36 {
 37     if(SegTree[rt].lazy!=0)
 38     {
 39         SegTree[rt<<1].lazy+=SegTree[rt].lazy;
 40         SegTree[rt<<1|1].lazy+=SegTree[rt].lazy;
 41         SegTree[rt<<1].sum+=(SegTree[rt<<1].r-SegTree[rt<<1].l+1)*SegTree[rt].lazy;
 42         SegTree[rt<<1|1].sum+=(SegTree[rt<<1|1].r-SegTree[rt<<1|1].l+1)*SegTree[rt].lazy;
 43         SegTree[rt].lazy=0;
 44     }
 45 }
 46 
 47 void Build(int l,int r,int rt)
 48 {
 49     SegTree[rt].l=l;
 50     SegTree[rt].r=r;
 51     SegTree[rt].lazy=0;//多样例时必须加 
 52     if(l==r)
 53     {
 54         scanf("%lld",&SegTree[rt].sum);
 55         return;
 56     }
 57     int mid=(l+r)>>1;
 58     Build(l,mid,rt<<1);
 59     Build(mid+1,r,rt<<1|1);
 60     PushUp(rt);
 61 }
 62 
 63 void Update(int L,int R,int add,int rt)
 64 {
 65     int l=SegTree[rt].l;
 66     int r=SegTree[rt].r;
 67     if(L<=l&&R>=r)
 68     {
 69         SegTree[rt].sum+=(SegTree[rt].r-SegTree[rt].l+1)*add;
 70         SegTree[rt].lazy+=add;
 71         return ;
 72     }
 73     PushDown(rt);//向下更新lazy 
 74     int mid=(l+r)>>1;
 75     if(L<=mid)
 76         Update(L,R,add,rt<<1);
 77     if(R>mid)
 78         Update(L,R,add,rt<<1|1);
 79     PushUp(rt);
 80 }
 81 
 82 LL Query(int L,int R,int rt)
 83 {
 84     int l=SegTree[rt].l;
 85     int r=SegTree[rt].r;
 86     if(L<=l&&R>=r)
 87     {
 88         return SegTree[rt].sum;
 89     }
 90     PushDown(rt);//向下更新lazy 
 91     int mid=(l+r)>>1;
 92     LL sum=0;
 93     if(L<=mid)
 94         sum+=Query(L,R,rt<<1);
 95     if(R>mid)
 96         sum+=Query(L,R,rt<<1|1);
 97     return sum;
 98 }
 99 
100 int main()
101 {
102     while(~scanf("%d %d",&n,&m))
103     {
104         Build(1,n,1);
105         for(int i=1;i<=m;i++)
106         {
107             char c[5];
108             int a,b;
109             scanf("%s %d %d",c,&a,&b);
110             if(c[0]=='Q')
111             {
112                 printf("%lld
",Query(a,b,1));
113             }
114             else if(c[0]=='C')
115             {
116                 int add;
117                 scanf("%d",&add);
118                 Update(a,b,add,1);
119             }
120         }
121     }
122     return 0;
123 }

View Code

更新：

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <iostream>
  4 #include <string>
  5 #include <math.h>
  6 #include <algorithm>
  7 #include <vector>
  8 #include <stack>
  9 #include <queue>
 10 #include <set>
 11 #include <map>
 12 #include <sstream>
 13 const int INF=0x3f3f3f3f;
 14 typedef long long LL;
 15 const double eps =1e-8;
 16 const int mod=1e9+7;
 17 const int maxn=1e5+10;
 18 using namespace std;
 19 
 20 struct node
 21 {
 22     int l;
 23     int r;
 24     LL val;
 25     LL lazy;
 26 }Tr[maxn<<2];
 27 
 28 void PU(int u)    //pushup 
 29 {
 30     Tr[u].val=Tr[u<<1].val+Tr[u<<1|1].val;
 31 }
 32 
 33 void PD(int u)    //pushdown
 34 {
 35     if(Tr[u].lazy)
 36     {
 37         Tr[u<<1].val+=Tr[u].lazy*(Tr[u<<1].r-Tr[u<<1].l+1);
 38         Tr[u<<1|1].val+=Tr[u].lazy*(Tr[u<<1|1].r-Tr[u<<1|1].l+1);
 39         Tr[u<<1].lazy+=Tr[u].lazy; Tr[u<<1|1].lazy+=Tr[u].lazy;
 40         Tr[u].lazy=0;
 41     }
 42 }
 43 
 44 void Build(int l,int r,int u)
 45 {
 46     Tr[u].l=l; Tr[u].r=r; Tr[u].lazy=0;
 47     if(l==r)
 48     {
 49         scanf("%lld",&Tr[u].val);
 50         return ;
 51     }
 52     int mid=(l+r)>>1;
 53     Build(l,mid,u<<1);
 54     Build(mid+1,r,u<<1|1);
 55     PU(u);
 56 }
 57 
 58 void Update(int L,int R,int u,int x)
 59 {
 60     int l=Tr[u].l; int r=Tr[u].r;
 61     if(L<=l&&R>=r)
 62     {
 63         Tr[u].val+=(Tr[u].r-Tr[u].l+1)*x;
 64         Tr[u].lazy+=x;
 65         return ;
 66     }
 67     PD(u);
 68     int mid=(l+r)>>1;
 69     if(L<=mid) Update(L,R,u<<1,x);
 70     if(R>mid) Update(L,R,u<<1|1,x);
 71     PU(u);
 72 }
 73 
 74 LL Query(int L,int R,int u)
 75 {
 76     int l=Tr[u].l; int r=Tr[u].r;
 77     if(L<=l&&R>=r)
 78     {
 79         return Tr[u].val;
 80     }
 81     PD(u);
 82     int mid=(l+r)>>1;
 83     LL sum=0;
 84     if(L<=mid) sum+=Query(L,R,u<<1);
 85     if(R>mid) sum+=Query(L,R,u<<1|1);
 86     return sum;
 87 }
 88 
 89 
 90 int main()
 91 {
 92     #ifdef DEBUG
 93     freopen("sample.txt","r",stdin);
 94     #endif
 95     
 96     int n,m;
 97     scanf("%d %d",&n,&m);
 98     Build(1,n,1);
 99     for(int i=1;i<=m;i++)
100     {
101         char op[5];
102         int a,b,c;
103         scanf("%s %d %d",op,&a,&b);
104         if(op[0]=='C')
105         {
106             scanf("%d",&c);
107             Update(a,b,1,c);
108         }
109         else if(op[0]=='Q')
110         {
111             printf("%lld
",Query(a,b,1));
112         }
113     }        
114     
115     return 0;
116 }

HDU-1698 Just a Hook（线段树区间修改）

http://acm.hdu.edu.cn/showproblem.php?pid=1698

Problem Description

In the game of DotA, Pudge’s meat hook is actually the most horrible thing for most of the heroes. The hook is made up of several consecutive metallic sticks which are of the same length.

Now Pudge wants to do some operations on the hook.
Let us number the consecutive metallic sticks of the hook from 1 to N. For each operation, Pudge can change the consecutive metallic sticks, numbered from X to Y, into cupreous sticks, silver sticks or golden sticks.
The total value of the hook is calculated as the sum of values of N metallic sticks. More precisely, the value for each kind of stick is calculated as follows:

For each cupreous stick, the value is 1.
For each silver stick, the value is 2.

For each golden stick, the value is 3.

Pudge wants to know the total value of the hook after performing the operations.
You may consider the original hook is made up of cupreous sticks.

Input

The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 10 cases.
For each case, the first line contains an integer N, 1<=N<=100,000, which is the number of the sticks of Pudge’s meat hook and the second line contains an integer Q, 0<=Q<=100,000, which is the number of the operations.
Next Q lines, each line contains three integers X, Y, 1<=X<=Y<=N, Z, 1<=Z<=3, which defines an operation: change the sticks numbered from X to Y into the metal kind Z, where Z=1 represents the cupreous kind, Z=2 represents the silver kind and Z=3 represents the golden kind.

Output

For each case, print a number in a line representing the total value of the hook after the operations. Use the format in the example.

Sample Input

Sample Output

Case 1: The total value of the hook is 24.

题意：

输入一个n表示一段长度为n的区间，有n个编号为1~n的点，初始值全部为1。有q个操作，每个操作有3个数：l，r，v(1<=v<=3)表示将区间l~r的所有元素修改为v。求经过q次修改后的整个区间的值之和。

该题是最典型的线段树区间修改问题，需要用到懒惰标记。

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <iostream>
  4 #include <string>
  5 #include <math.h>
  6 #include <algorithm>
  7 #include <vector>
  8 #include <queue>
  9 #include <set>
 10 #include <map>
 11 #include <math.h>
 12 const int INF=0x3f3f3f3f;
 13 typedef long long LL;
 14 const int mod=1e9+7;
 15 //const double PI=acos(-1);
 16 const int maxn=1e5+10;
 17 using namespace std;
 18 //ios::sync_with_stdio(false);
 19 //    cin.tie(NULL);
 20 
 21 int n,m;
 22 struct node
 23 {
 24     int l;
 25     int r;
 26     int lazy;
 27     int sum;
 28 }SegTree[maxn<<2];
 29 
 30 void PushUp(int rt)
 31 {
 32     SegTree[rt].sum=SegTree[rt<<1].sum+SegTree[rt<<1|1].sum;
 33 }
 34 
 35 void PushDown(int rt)
 36 {
 37     if(SegTree[rt].lazy!=0)
 38     {
 39         //该题用=而不是+= 
 40         SegTree[rt<<1].lazy=SegTree[rt].lazy;
 41         SegTree[rt<<1|1].lazy=SegTree[rt].lazy;
 42         SegTree[rt<<1].sum=(SegTree[rt<<1].r-SegTree[rt<<1].l+1)*SegTree[rt].lazy;
 43         SegTree[rt<<1|1].sum=(SegTree[rt<<1|1].r-SegTree[rt<<1|1].l+1)*SegTree[rt].lazy;
 44         SegTree[rt].lazy=0;
 45     }
 46 }
 47 
 48 void Build(int l,int r,int rt)
 49 {
 50     SegTree[rt].l=l;
 51     SegTree[rt].r=r;
 52     SegTree[rt].lazy=0;//多样例时必须加 
 53     if(l==r)
 54     {
 55         SegTree[rt].sum=1;//该题不用输入，直接初始为1 
 56         return;
 57     }
 58     int mid=(l+r)>>1;
 59     Build(l,mid,rt<<1);
 60     Build(mid+1,r,rt<<1|1);
 61     PushUp(rt);
 62 }
 63 
 64 void Update(int L,int R,int C,int rt)
 65 {
 66     int l=SegTree[rt].l;
 67     int r=SegTree[rt].r;
 68     if(L<=l&&R>=r)
 69     {
 70         SegTree[rt].sum=(SegTree[rt].r-SegTree[rt].l+1)*C;//该题用=而不是+= 
 71         SegTree[rt].lazy=C;//该题用=而不是+= 
 72         return ;
 73     }
 74     PushDown(rt);//向下更新lazy 
 75     int mid=(l+r)>>1;
 76     if(L<=mid)
 77         Update(L,R,C,rt<<1);
 78     if(R>mid)
 79         Update(L,R,C,rt<<1|1);
 80     PushUp(rt);
 81 }
 82 
 83 int Query(int L,int R,int rt)
 84 {
 85     int l=SegTree[rt].l;
 86     int r=SegTree[rt].r;
 87     if(L<=l&&R>=r)
 88     {
 89         return SegTree[rt].sum;
 90     }
 91     PushDown(rt);//向下更新lazy
 92     int mid=(l+r)>>1;
 93     int sum=0;
 94     if(L<=mid)
 95         sum+=Query(L,R,rt<<1);
 96     if(R>mid)
 97         sum+=Query(L,R,rt<<1|1);
 98     return sum;
 99 }
100 
101 int main()
102 {
103     int T;    
104     scanf("%d",&T);
105     for(int k=1;k<=T;k++)
106     {
107         scanf("%d %d",&n,&m);
108         Build(1,n,1);
109         for(int i=1;i<=m;i++)
110         {
111             int a,b,c;
112             scanf("%d %d %d",&a,&b,&c);
113             Update(a,b,c,1);
114         }
115         printf("Case %d: The total value of the hook is %d.
",k,SegTree[1].sum);
116     }
117     return 0;
118 }

更新版本

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <iostream>
  4 #include <string>
  5 #include <math.h>
  6 #include <algorithm>
  7 #include <vector>
  8 #include <stack>
  9 #include <queue>
 10 #include <set>
 11 #include <map>
 12 #include <sstream>
 13 typedef long long LL;
 14 const int INF = 0x3f3f3f3f;//0x7fffffff;
 15 const double eps = 1e-8;
 16 const int mod = 1e9+7;
 17 const int maxn = 1e5+10;
 18 using namespace std;
 19 
 20 struct node
 21 {
 22     int l, r;
 23     LL val;
 24     LL lazy;
 25 }Tr[maxn<<2];
 26 
 27 void PU(int u)    //Pushup 
 28 {
 29     Tr[u].val = Tr[u<<1].val+Tr[u<<1|1].val;
 30 }
 31 
 32 void PD(int u)    //Pushdown
 33 {
 34     if(Tr[u].lazy)
 35     {   //该题用=而不是+=
 36         Tr[u<<1].val = Tr[u].lazy*(Tr[u<<1].r-Tr[u<<1].l+1);
 37         Tr[u<<1|1].val = Tr[u].lazy*(Tr[u<<1|1].r-Tr[u<<1|1].l+1);
 38         Tr[u<<1].lazy = Tr[u].lazy; Tr[u<<1|1].lazy = Tr[u].lazy;
 39         Tr[u].lazy = 0;
 40     }
 41 }
 42 
 43 void Build(int l, int r, int u) //建树 
 44 {
 45     Tr[u].l = l; Tr[u].r = r; Tr[u].lazy = 0; //多样例时必须加lazy=0
 46     if(l == r)
 47     {
 48         Tr[u].val = 1;//该题不用输入，直接初始为1
 49         return ;
 50     }
 51     int mid = (l+r)>>1;
 52     Build(l, mid, u<<1);
 53     Build(mid+1, r, u<<1|1);
 54     PU(u);
 55 }
 56 
 57 void Update(int L, int R, int u, LL x)  //区间更新
 58 {
 59     int l = Tr[u].l; int r = Tr[u].r;
 60     if(L<=l && R>=r)
 61     {
 62         Tr[u].val = x*(r-l+1);  //该题用=而不是+=
 63         Tr[u].lazy = x;  //该题用=而不是+=
 64         return ;
 65     }
 66     PD(u);  //向下更新lazy
 67     int mid = (l+r)>>1;
 68     if(L <= mid) Update(L, R, u<<1, x);
 69     if(R > mid) Update(L, R, u<<1|1, x);
 70     PU(u);
 71 }
 72 
 73 LL Query(int L, int R, int u)    //区间查询 
 74 {
 75     int l = Tr[u].l; int r=Tr[u].r;
 76     if(L<=l && R>=r)
 77     {
 78         return Tr[u].val;
 79     }
 80     PD(u);  //向下更新lazy
 81     int mid = (l+r)>>1;
 82     LL sum = 0;
 83     if(L <= mid) sum += Query(L, R, u<<1);
 84     if(R > mid) sum += Query(L, R, u<<1|1);
 85     return sum;
 86 }
 87 
 88 int main()
 89 {
 90     #ifdef DEBUG
 91     freopen("sample.txt","r",stdin);  //freopen("data.out", "w", stdout);
 92     #endif
 93 
 94     int T;
 95     scanf("%d",&T);
 96     for(int cs=1;cs<=T;cs++)
 97     {
 98         int n,m;
 99         scanf("%d %d",&n,&m);
100         Build(1,n,1);
101         for(int i=1;i<=m;i++)
102         {
103             int a,b,c;
104             scanf("%d %d %d",&a,&b,&c);
105             Update(a,b,1,c);
106         }
107         printf("Case %d: The total value of the hook is %lld.
",cs,Tr[1].val);
108     }
109 
110     return 0;
111 }

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原文地址：https://www.cnblogs.com/jiamian/p/11392142.html