hdu-1892 See you~---二维树状数组运用

Posted 2020-11-01 努力努力再努力x

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1892

题目大意：

题目大意：有很多方格，每个方格对应的坐标为（I,J),刚开始时每个格子里有1本书，然后让你统计一片区域有多少本书，还可以增加书和减少，移动书。

解题思路：

直接二维数组数组模拟

注意：

每个下标+1，从(1, 1)开始

求区域和的时候给出的x1 y1 和x2 y2不是标准的正对角线，需要转化

  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 #define MEM(a, x) memset(a, x, sizeof(a));
 19 using namespace std;
 20 typedef long long ll;
 21 typedef unsigned long long ull;
 22 const int maxn = 1010;
 23 const double eps = 1e-10;
 24 const int INF = 1 << 30;
 25 const int dir[4][2] = {1,0,0,1,0,-1,-1,0};
 26 const double pi = 3.1415926535898;
 27 int T, n, m, cases;
 28 int tree[maxn][maxn], a[maxn][maxn];
 29 int lowbit(int x)
 30 {
 31     return x&(-x);
 32 }
 33 int sum(int x, int y)
 34 {
 35     int ans = 0;
 36     for(int i = x; i > 0; i -= lowbit(i))
 37         for(int j = y; j > 0; j -= lowbit(j))
 38         ans += tree[i][j];
 39     return ans;
 40 }
 41 void add(int x, int y, int d)
 42 {
 43     for(int i = x; i < maxn; i += lowbit(i))
 44     {
 45         for(int j = y; j < maxn; j += lowbit(j))
 46         {
 47             tree[i][j] += d;
 48         }
 49     }
 50 }
 51 int main()
 52 {
 53     scanf("%d", &T);
 54     char s[3];
 55     int x, y, x1, y1, x2, y2, d;
 56     while(T--)
 57     {
 58         scanf("%d", &n);
 59         MEM(tree, 0);
 60         for(int i = 1; i < maxn; i++)
 61         {
 62             for(int j = 1; j < maxn; j++)
 63             {
 64                 add(i, j, 1);
 65                 a[i][j] = 1;
 66             }
 67         }
 68         printf("Case %d:\n", ++cases);
 69         while(n--)
 70         {
 71             scanf("%s", s);
 72             if(s[0] == ‘S‘)
 73             {
 74                 scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
 75                 ///不是标准的x1<x2, y1<y2，要对其进行转化
 76                 int a, b, c, d;
 77                 a = min(x1, x2);a++;
 78                 b = min(y1, y2);b++;
 79                 c = max(x1, x2);c++;
 80                 d = max(y1, y2);d++;
 81                 printf("%d\n", sum(c, d) + sum(a - 1, b - 1) - sum(a - 1, d) - sum(c, b - 1));
 82             }
 83             else if(s[0] == ‘A‘)
 84             {
 85                 scanf("%d%d%d", &x, &y, &d);
 86                 x++, y++;
 87                 add(x, y, d);
 88                 a[x][y] += d;
 89             }
 90             else if(s[0] == ‘D‘)
 91             {
 92                 scanf("%d%d%d", &x, &y, &d);
 93                 x++, y++;
 94                 if(d > a[x][y])d = a[x][y];
 95                 add(x, y, -d);
 96                 a[x][y] -= d;
 97             }
 98             else
 99             {
100                 scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &d);
101                 x1++, y1++, x2++, y2++;
102                 if(d > a[x1][y1])d = a[x1][y1];
103                 add(x1, y1, -d);
104                 add(x2, y2, d);
105                 a[x1][y1] -= d;
106                 a[x2][y2] += d;
107             }
108         }
109     }
110     return 0;
111 }