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A Simple Problem with Integers POJ - 3468 (线段树)

时间:2019-09-04 13:30:44      阅读:78      评论:0      收藏:0      [点我收藏+]

标签:scanf   stream   with   imp   else   list   tor   ret   oid   

技术图片

 

 

思路:线段树,区间更新,区间查找

  1 #include<iostream>
  2 #include<vector>
  3 #include<string>
  4 #include<cmath>
  5 #include<set>
  6 #include<algorithm>
  7 #include<cstdio>
  8 #include<map>
  9 #include<cstring>
 10 #include<list>
 11 
 12 #define MAXSIZE 100010
 13 
 14 using namespace std;
 15 
 16 long long tree[4*MAXSIZE];
 17 long long lz[4*MAXSIZE];
 18 int N, Q;
 19 
 20 void init()
 21 {
 22     memset(tree, 0, sizeof(tree));
 23     memset(lz, 0, sizeof(lz));
 24 }
 25 
 26 void build(int node, int l, int r)
 27 {
 28     if(l == r)
 29     {
 30         scanf("%lld", &tree[node]);
 31         return;
 32     }
 33     
 34     int mid = (l+r)/2;
 35     build(node*2, l, mid);
 36     build(node*2+1, mid+1, r);
 37     
 38     tree[node] = tree[node*2] + tree[node*2+1];
 39 }
 40 
 41 void push_down(int node, int l, int r)
 42 {
 43     if(lz[node])
 44     {
 45         int mid = (l+r)/2;
 46         lz[node*2] += lz[node];
 47         lz[node*2+1] += lz[node];
 48         tree[node*2] += (mid-l+1)*lz[node];
 49         tree[node*2+1] += (r-mid)*lz[node];
 50         lz[node] = 0;
 51     }
 52 }
 53 
 54 void update_range(int node, int l, int r, int L, int R, int add)
 55 {
 56     if(l <= L && r >= R)
 57     {
 58         lz[node] += add;
 59         tree[node] += (R-L+1)*add;
 60         return;
 61     }
 62     push_down(node, L, R);
 63     int mid = (L+R)/2;
 64     if(mid >= l)
 65         update_range(node*2, l, r, L, mid, add);
 66     if(mid < r)
 67         update_range(node*2+1, l, r, mid+1, R, add);
 68     tree[node] = tree[node*2] + tree[node*2+1];
 69 }
 70 
 71 long long query_range(int node, int l, int r, int L, int R)
 72 {
 73     if(l <= L && r >= R)
 74     {
 75         return tree[node];
 76     }
 77     push_down(node, L, R);
 78     int mid = (L+R)/2;
 79     long long sum = 0;
 80     if(mid >= l)
 81         sum += query_range(node*2, l, r, L, mid);
 82     if(mid < r)
 83         sum += query_range(node*2+1, l, r, mid+1, R);
 84     
 85     return sum;
 86 }
 87 
 88 
 89 
 90 int main()
 91 {    
 92     scanf("%d%d", &N, &Q);
 93     init();
 94     build(1, 1, N);
 95     while(Q--)
 96     {
 97         char ch;
 98         scanf(" %c", &ch);
 99         if(ch == Q)
100         {
101             int a, b;
102             scanf("%d%d", &a, &b);
103             printf("%lld\n", query_range(1, a, b, 1, N));
104         }
105         else if(ch == C)
106         {
107             int a, b, c;
108             scanf("%d%d%d", &a, &b, &c);
109             update_range(1, a, b, 1, N, c);
110         }
111     }
112     
113 
114     return 0;
115 }

 

A Simple Problem with Integers POJ - 3468 (线段树)

标签:scanf   stream   with   imp   else   list   tor   ret   oid   

原文地址:https://www.cnblogs.com/FengZeng666/p/11458279.html

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