标签:targe mat mem AC stream sizeof poj nbsp queue
题目链接:
https://vjudge.net/problem/POJ-1195
题目大意:
直接维护二维树状数组
注意横纵坐标全部需要加1,因为树状数组从(1,1)开始
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<string> 6 #include<cmath> 7 #include<set> 8 #include<queue> 9 #include<map> 10 #include<stack> 11 #include<vector> 12 #include<list> 13 #include<deque> 14 #include<sstream> 15 #include<cctype> 16 #define REP(i, n) for(int i = 0; i < (n); i++) 17 #define FOR(i, s, t) for(int i = (s); i < (t); i++) 18 using namespace std; 19 typedef long long ll; 20 int T, n, m, cases; 21 int num[1048][1048]; 22 int lowbit(int x) 23 { 24 return x & (-x); 25 } 26 ///修改num[x][y] = d 27 void add(int x, int y, int d) 28 { 29 for(int i = x; i <= n; i += lowbit(i)) 30 { 31 for(int j = y; j <= n; j += lowbit(j)) 32 num[i][j] += d; 33 } 34 } 35 ///计算顶点为(1, 1)(x, y)的数总和 36 int sum(int x, int y) 37 { 38 int ans = 0; 39 for(int i = x; i > 0; i -= lowbit(i)) 40 { 41 for(int j = y; j > 0; j -= lowbit(j)) 42 ans += num[i][j]; 43 } 44 return ans; 45 } 46 int main() 47 { 48 int t, x1, y1, x2, y2, d; 49 while(cin >> t >> n) 50 { 51 memset(num, 0, sizeof(num)); 52 while(cin >> t) 53 { 54 if(t == 3)break; 55 else if(t == 1) 56 { 57 scanf("%d%d%d", &x1, &y1, &d); 58 x1++, y1++; 59 add(x1, y1, d); 60 } 61 else if(t == 2) 62 { 63 scanf("%d%d%d%d", &x1, &y1, &x2, &y2); 64 x1++, y1++, x2++, y2++; 65 cout<<(sum(x2, y2) - sum(x2, y1 - 1) - sum(x1 - 1, y2) + sum(x1 - 1, y1 - 1))<<endl; 66 } 67 } 68 } 69 return 0; 70 }
POJ-1195 Mobile phones---裸的二维树状数组(注意下标从1,1开始)
标签:targe mat mem AC stream sizeof poj nbsp queue
原文地址:https://www.cnblogs.com/fzl194/p/8934156.html