See you~
Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4768 Accepted Submission(s): 1521
Problem Description
Now
I am leaving hust acm. In the past two and half years, I learned so
many knowledge about Algorithm and Programming, and I met so many good
friends. I want to say sorry to Mr, Yin, I must leave now ~~>.<~~.
I am very sorry, we could not advanced to the World Finals last year.
When
coming into our training room, a lot of books are in my eyes. And every
time the books are moving from one place to another one. Now give you
the position of the books at the early of the day. And the moving
information of the books the day, your work is to tell me how many books
are stayed in some rectangles.
To make the problem easier, we
divide the room into different grids and a book can only stayed in one
grid. The length and the width of the room are less than 1000. I can
move one book from one position to another position, take away one book
from a position or bring in one book and put it on one position.
Input
In
the first line of the input file there is an Integer T(1<=T<=10),
which means the number of test cases in the input file. Then N test
cases are followed.
For each test case, in the first line there is
an Integer Q(1<Q<=100,000), means the queries of the case. Then
followed by Q queries.
There are 4 kind of queries, sum, add, delete and move.
For example:
S
x1 y1 x2 y2 means you should tell me the total books of the rectangle
used (x1,y1)-(x2,y2) as the diagonal, including the two points.
A x1 y1 n1 means I put n1 books on the position (x1,y1)
D x1 y1 n1 means I move away n1 books on the position (x1,y1), if less than n1 books at that position, move away all of them.
M
x1 y1 x2 y2 n1 means you move n1 books from (x1,y1) to (x2,y2), if less
than n1 books at that position, move away all of them.
Make sure that at first, there is one book on every grid and 0<=x1,y1,x2,y2<=1000,1<=n1<=100.
Output
At the beginning of each case, output "Case X:" where X is the index of the test case, then followed by the "S" queries.
For each "S" query, just print out the total number of books in that area.
Sample Input
2
3
S 1 1 1 1
A 1 1 2
S 1 1 1 1
3
S 1 1 1 1
A 1 1 2
S 1 1 1 2
Sample Output
Case 1:
1
3
Case 2:
1
4
Author
Sempr|CrazyBird|hust07p43
思路：二维树状数组；
套个二维的模板就行，注意给的两个点的大小关系；
1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<stdlib.h>
5 #include<queue>
6 #include<string.h>
7 #include<map>
8 #include<vector>
9 #include<queue>
10 using namespace std;
11 typedef long long LL;
12 int ma[1005][1005];
13 int bit[1005][1005];
14 int cit[1005][1005];
15 int lowbit(int x);
16 void add(int x,int y,int c,int v);
17 int ask(int x,int y);
18 int main(void)
19 {
20 int n;
21 scanf("%d",&n);
22 int __ca = 0;
23 int i,j;
24 for(i = 1; i <= 1001; i++)
25 {
26 for(j = 1; j <= 1001; j++)
27 {
28 ma[i][j] = 1;
29 add(i,j,1,0);
30 }
31 }
32 while(n--)
33 {
34 memset(bit,0,sizeof(bit));
35 int m;
36 for(i = 0; i <= 1001; i++)
37 {
38 for(j = 0; j <= 1001; j++)
39 {
40 bit[i][j] = cit[i][j];
41 }
42 }
43 for(i = 0; i <= 1001; i++)
44 {
45 for(j = 0; j <= 1001; j++)
46 {
47 ma[i][j] = 1;
48 }
49 }
50 char ans[10];
51 scanf("%d",&m);
52 printf("Case %d:\n",++__ca);
53 while(m--)
54 {
55 scanf("%s",ans);
56 int x,y,x1,y1;
57 if(ans[0]==‘S‘)
58 {
59 scanf("%d %d %d %d",&x,&y,&x1,&y1);
60 if(x > x1)
61 swap(x,x1),swap(y,y1);
62 if(y > y1)
63 {
64 swap(y,y1);
65 }
66 x++;
67 y++;
68 x1++;
69 y1++;
70 int sum = ask(x1,y1);
71 sum -= ask(x-1,y1);
72 sum -= ask(x1,y-1);
73 sum += ask(x-1,y-1);
74 printf("%d\n",sum);
75 }
76 else if(ans[0] == ‘A‘)
77 {
78 int c;
79 scanf("%d %d %d",&x,&y,&c);
80 x++;
81 y++;
82 ma[x][y] += c;
83 add(x,y,c,1);
84 }
85 else if(ans[0] == ‘M‘)
86 {
87 int c;
88 scanf("%d %d %d %d %d",&x,&y,&x1,&y1,&c);
89 x++;
90 y++;
91 x1++;
92 y1++;
93 if(ma[x][y] < c)
94 c = ma[x][y];
95 ma[x][y] -= c;
96 ma[x1][y1]+=c;
97 add(x,y,-c,1);
98 add(x1,y1,c,1);
99 }
100 else
101 {
102 int c;
103 scanf("%d %d %d",&x,&y,&c);
104 x++;
105 y++;
106 if(ma[x][y] < c)
107 c = ma[x][y];
108 ma[x][y] -= c;
109 add(x,y,-c,1);
110 }
111 }
112 }
113 return 0;
114 }
115 int lowbit(int x)
116 {
117 return x&(-x);
118 }
119 void add(int x,int y,int c,int v)
120 {
121 int i,j;
122 for(i = x; i <= 1001; i += lowbit(i))
123 {
124 for(j = y; j <= 1001; j += lowbit(j))
125 {
126 if(v)
127 bit[i][j] += c;
128 else cit[i][j]+=c;
129 }
130 }
131 }
132 int ask(int x,int y)
133 {
134 int i,j;
135 int sum = 0;
136 for(i = x; i > 0; i -= lowbit(i))
137 {
138 for(j = y; j > 0; j -= lowbit(j))
139 {
140 sum += bit[i][j];
141 }
142 }
143 return sum;
144 }