标签:names using 操作 string sizeof include bre 树状 class
题意:
对一个矩阵有2种操作:
1.把某个元素设为x。
2.查询以(x1,y1)为左上角 以(x2,y2)为右上角的矩阵中的数字的和。
思路:
二维树状数组入门题,同时对横坐标和纵坐标做前缀和就行了。
代码:
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 const int N = 1030; 6 int a[N][N]; 7 int c[N][N]; 8 int n; 9 int lowbit(int x) 10 { 11 return x&(-x); 12 } 13 void add(int x,int y,int ch) 14 { 15 for (int i = x;i <= n;i += lowbit(i)) 16 { 17 for (int j = y;j <= n;j += lowbit(j)) 18 { 19 c[i][j] += ch; 20 } 21 } 22 } 23 int getsum(int x,int y) 24 { 25 int ans = 0; 26 for (int i = x;i > 0;i -= lowbit(i)) 27 { 28 for (int j = y;j > 0;j -= lowbit(j)) 29 { 30 ans += c[i][j]; 31 } 32 } 33 return ans; 34 } 35 int main() 36 { 37 int T; 38 scanf("%d",&T); 39 while (T--) 40 { 41 memset(c,0,sizeof(c)); 42 memset(a,0,sizeof(a)); 43 scanf("%d",&n); 44 char s[5]; 45 while (scanf("%s",s) != EOF) 46 { 47 if (s[1] == ‘N‘) break; 48 if (s[1] == ‘E‘) 49 { 50 int x,y,num; 51 scanf("%d%d%d",&x,&y,&num); 52 x++,y++; 53 int ch = num - a[x][y]; 54 a[x][y] = num; 55 add(x,y,ch); 56 } 57 if (s[1] == ‘U‘) 58 { 59 int x,y,p,q; 60 scanf("%d%d%d%d",&x,&y,&p,&q); 61 x++,y++,p++,q++; 62 int ans = 0; 63 ans += getsum(p,q); 64 ans -= getsum(x-1,q); 65 ans -= getsum(p,y-1); 66 ans += getsum(x-1,y-1); 67 printf("%d\n",ans); 68 } 69 } 70 } 71 return 0; 72 }
标签:names using 操作 string sizeof include bre 树状 class
原文地址:https://www.cnblogs.com/kickit/p/9073721.html