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  • 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
    ", 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 }
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  • 原文地址:https://www.cnblogs.com/Surprisezang/p/9075275.html
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