poj3468 A Simple Problem with Integers（线段树区间更新）

https://vjudge.net/problem/POJ-3468

线段树区间更新（lazy数组）模板题

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<map>
 8 #define lson l, m, rt<<1
 9 #define rson m+1, r, rt<<1|1
10 #define INF 0x3f3f3f3f
11 typedef long long ll;
12 using namespace std;
13 ll sum[100010<<2], lazy[100010<<2];
14 void PushUP(int rt)
15 {
16     sum[rt] = sum[rt<<1]+sum[rt<<1|1];
17 }
18 void PushDown(int rt, int m)
19 {
20     if(lazy[rt]){//若有标记下移 
21         lazy[rt<<1] += lazy[rt];
22         lazy[rt<<1|1] += lazy[rt]; 
23         sum[rt<<1] += (m-(m>>1))*lazy[rt];
24         sum[rt<<1|1] += (m>>1)*lazy[rt];
25         lazy[rt] = 0;//取消本层标记 
26     }
27 } 
28 void build(int l, int r, int rt)
29 {
30     lazy[rt] = 0;//一定写在if外初始化 
31     if(l == r){
32         scanf("%lld", &sum[rt]);
33         return ;
34     }
35     int m = (l+r)>>1;
36     build(lson);
37     build(rson);
38     PushUP(rt);
39 } 
40 ll Query(int L, int R, int l, int r, int rt)
41 {
42     if(L<=l&&R>=r){
43         return sum[rt];
44     }
45     PushDown(rt, r-l+1);//向下更新 
46     int m = (l+r)>>1;
47     ll ans=0;
48     if(L <= m) ans += Query(L, R, lson);
49     if(R > m) ans += Query(L, R, rson);
50     return ans;
51 }
52 void Update(int L, int R, ll add, int l, int r, int rt)
53 {
54     if(L<=l&&R>=r){
55         sum[rt] += (r-l+1)*add;
56         lazy[rt] += add;
57         return ;
58     }
59     PushDown(rt, r-l+1);//向下而更新 
60     int m=(l+r)>>1;
61     if(L <= m) Update(L, R, add, lson);
62     if(R > m) Update(L, R, add, rson);
63     PushUP(rt);
64 }
65 int main()
66 {
67     int n, m, a, b;
68     ll c;
69     char op[3];
70     scanf("%d%d", &n, &m);
71     build(1, n, 1);
72     for(int i = 0; i < m; i++){
73         scanf("%s", &op);
74         if(op[0] == 'Q'){
75             scanf("%d%d", &a, &b);
76             printf("%lld\n", Query(a, b, 1, n, 1));
77         }
78         else{
79             scanf("%d%d%lld", &a, &b, &c);
80             Update(a, b, c, 1, n, 1);
81         }
82     } 
83     return 0;
84 }