poj 1195（二維線段樹||二維樹狀數組）

Mobile phones

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 17496		Accepted: 8089

Description Suppose that the fourth generation mobile phone base stations in the Tampere area operate as follows. The area is divided into squares. The squares form an S * S matrix with the rows and columns numbered from 0 to S-1. Each square contains a base station. The number of active mobile phones inside a square can change because a phone is moved from a square to another or a phone is switched on or off. At times, each base station reports the change in the number of active phones to the main base station along with the row and the column of the matrix. 

Write a program, which receives these reports and answers queries about the current total number of active mobile phones in any rectangle-shaped area. 

Input The input is read from standard input as integers and the answers to the queries are written to standard output as integers. The input is encoded as follows. Each input comes on a separate line, and consists of one instruction integer and a number of parameter integers according to the following table. 

The values will always be in range, so there is no need to check them. In particular, if A is negative, it can be assumed that it will not reduce the square value below zero. The indexing starts at 0, e.g. for a table of size 4 * 4, we have 0 <= X <= 3 and 0 <= Y <= 3. 

Table size: 1 * 1 <= S * S <= 1024 * 1024 
Cell value V at any time: 0 <= V <= 32767 
Update amount: -32768 <= A <= 32767 
No of instructions in input: 3 <= U <= 60002 
Maximum number of phones in the whole table: M= 2^30 

Output Your program should not answer anything to lines with an instruction other than 2. If the instruction is 2, then your program is expected to answer the query by writing the answer as a single line containing a single integer to standard output.

Sample Input

0 41 1 2 32 0 0 2 2 1 1 1 21 1 2 -12 1 1 2 3 3

Sample Output

34

Source IOI 2001

題意：

一個由數字構成的大矩陣，開始是全0,能進行兩種操作
1) 對矩陣裡的某個數加上一個整數（可正可負）
2) 查詢某個子矩陣裡所有數位和
要求對每次查詢，輸出結果

題解：模板題，使用二維線段樹或者二維樹狀數組數組

以下給出2種做法：

1：二維樹狀數組

#include<cstdio>  #include<cstring>  #include<cstdlib>  #include<cmath>  #include<iostream>  #include<algorithm>  #include<vector>  #include<map>  #include<set>  #include<queue>  #include<string>  #include<bitset>  #include<utility>  #include<functional>  #include<iomanip>  #include<sstream>  #include<ctime>  using namespace std;#define N int(3e3)  #define inf int(0x3f3f3f3f)  #define mod int(1e9+7)  typedef long long LL;#ifdef CDZSC  #define debug(...) fprintf(stderr, __VA_ARGS__)  #else  #define debug(...)   #endif  int n,c[N][N];int lowbit(int x){return x&(-x);}int query_sum(int x, int y){int res = 0;for (int i = x; i > 0; i -= lowbit(i)){for (int j = y; j > 0; j -= lowbit(j)){res += c[i][j];}}return res;}void add(int x, int y, int val){for (int i = x; i <= n; i += lowbit(i)){for (int j = y; j <= n; j += lowbit(j)){c[i][j] += val;}}}int main(){#ifdef CDZSC  freopen("i.txt", "r", stdin);//freopen("o.txt","w",stdout);  int _time_jc = clock();#endif  int s, x, y, a, L, b, R, t;while (~scanf("%d%d", &s, &n)){memset(c, 0, sizeof(c));while (~scanf("%d", &s)){if (s == 3)break;if (s == 1){scanf("%d%d%d", &x, &y, &a);add(x + 1, y + 1, a);}else{scanf("%d%d%d%d", &L, &b, &R, &t);L++; b++; R++; t++;printf("%d\n", query_sum(R, t) - query_sum(R, b - 1) - query_sum(L - 1, t) + query_sum(L - 1, b - 1));}}}#ifdef CDZSC  debug("time: %d\n", int(clock() - _time_jc));#endif  return 0;}

2：二維線段樹

#include<cstdio>  #include<cstring>  #include<cstdlib>  #include<cmath>  #include<iostream>  #include<algorithm>  #include<vector>  #include<map>  #include<set>  #include<queue>  #include<string>  #include<bitset>  #include<utility>  #include<functional>  #include<iomanip>  #include<sstream>  #include<ctime>  using namespace std;#define N int(1e3+50)  #define inf int(0x3f3f3f3f)  #define mod int(1e9+7)  typedef long long LL;#ifdef CDZSC  #define debug(...) fprintf(stderr, __VA_ARGS__)  #else  #define debug(...)   #endif  int S,tree[N<<2][N<<2];void update_x(int rooty, int rootx, int L, int R, int x, int a) //tree[rooty][rootx]對應的矩陣x方向上範圍是[L,R]{ tree[rooty][rootx] += a; if( L == R ) return; int mid = (L + R )/2; if( x <= mid ) update_x(rooty,( rootx << 1) + 1, L ,mid, x, a);else update_x(rooty,( rootx << 1) + 2, mid + 1,R, x, a);}void update_y(int rooty, int L, int R, int y, int x, int a) //tree[rooty][rootx]對應的矩陣y方向上範圍是[L,R] { update_x( rooty,0, 1, S, x,a); if( L == R) return; int mid = (L + R )/2;if( y <= mid ) update_y( ( rooty << 1) + 1, L, mid,y, x, a); elseupdate_y( ( rooty << 1) + 2, mid+1, R, y, x, a); }int query_x(int rooty, int rootx, int L, int R, int x1, int x2) { if (L == x1 && R == x2)return tree[rooty][rootx]; int mid = (L + R) / 2; if (x2 <= mid)return query_x(rooty, (rootx << 1) + 1, L, mid, x1, x2); else if (x1 > mid)return query_x(rooty, (rootx << 1) + 2, mid + 1, R, x1, x2); elsereturn query_x(rooty, (rootx << 1) + 1, L, mid, x1, mid) + query_x(rooty, (rootx << 1) + 2, mid + 1, R, mid + 1, x2); }int query_y(int rooty, int L, int R, int y1, int y2, int x1, int x2){if (L == y1 && R == y2)return query_x(rooty, 0, 1, S, x1, x2);int mid = (L + R) / 2;if (y2 <= mid)return query_y((rooty << 1) + 1, L,mid, y1, y2, x1, x2);if (y1 > mid)return query_y((rooty << 1) + 2,mid + 1, R, y1, y2, x1, x2);elsereturn query_y((rooty << 1) + 1, L,mid, y1, mid, x1, x2) +query_y((rooty << 1) + 2,mid + 1, R, mid + 1, y2, x1, x2);}int main(){#ifdef CDZSC  freopen("i.txt", "r", stdin);//freopen("o.txt","w",stdout);  int _time_jc = clock();#endif  int n, s, x, y, a, L, b, R, t;while (~scanf("%d%d", &s, &S)){while (~scanf("%d", &s)){if (s == 3)break;if (s == 1){scanf("%d%d%d", &x, &y, &a);update_y(0, 1, S, y + 1, x + 1, a);}else{scanf("%d%d%d%d", &L, &b, &R, &t);L++; b++; R++; t++;printf("%d\n", query_y(0, 1, S,b, t, L, R));}}}#ifdef CDZSC  debug("time: %d\n", int(clock() - _time_jc));#endif  return 0;}