E

题意给出一段初始化全为1的区间后面可以一段一段更改成 1 或 2 或3 问最后整段区间的和是多少
思路：标准线段树区间和模版题
 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<set>
 4 #include<vector>
 5 #include<cstring>
 6 #include<iostream>
 7 using namespace std;
 8 const int maxn=1e5+7;
 9 struct Node{
10     int l,r,sum;
11     int lazy;
12     void update(int val){
13         sum=(r-l+1)*val;
14         lazy=val;
15 
16     }
17 }tree[maxn*4];
18 void push_up(int x){
19     tree[x].sum=tree[x<<1].sum+tree[x<<1|1].sum;
20 }
21 void push_down(int x){
22     if(tree[x].lazy!=0){
23         tree[x<<1].update(tree[x].lazy);
24         tree[x<<1|1].update(tree[x].lazy);
25         tree[x].lazy=0;
26     }
27 }
28 void build(int x,int l,int r){
29     tree[x].l=l,tree[x].r=r;
30     tree[x].sum=tree[x].lazy=0;
31     if(l==r){
32         tree[x].sum=1;
33     }
34     else {
35         int mid=l+r>>1;
36         build(x<<1,l,mid);
37         build(x<<1|1,mid+1,r);
38         push_down(x);
39     }
40 }
41 void update(int x,int l,int r,int c){
42     int L=tree[x].l,R=tree[x].r;
43     if(R<=r&&L>=l){
44         tree[x].update(c);
45     }
46     else {
47         int mid=L+R>>1;
48         push_down(x);
49         if(mid>=l)update(x<<1,l,r,c);
50         if(mid<r)update(x<<1|1,l,r,c);
51         push_up(x);
52     }
53 }
54 long long query(int x,int l,int r){
55     int     L=tree[x].l,R=tree[x].r;
56     if(R<=r&&L>=l){
57         return tree[x].sum;
58     }
59     else {
60         int mid=L+R>>1;
61         long long ans=0;
62         //push_down(x);
63         if(mid>=l)ans+=query(x<<1,l,r);
64         if(mid<r)ans+=query(x<<1|1,l,r);
65         return ans;
66     }
67 }
68 int main(){
69     int t,kase=1;
70     scanf("%d",&t);
71     while(t--){
72         int n,q;
73         scanf("%d%d",&n,&q);
74         build(1,1,n);
75         for(int i=0;i<q;i++){
76             int x,y,z;
77             scanf("%d%d%d",&x,&y,&z);
78             update(1,x,y,z);
79         }
80         printf("Case %d: The total value of the hook is %lld.
",kase++,query(1,1,n));
81     }
82     return 0;
83 }