HDOJ 1892 HDU 1892 See you~ ACM 1892 IN HDU

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋 代码See you~

Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 921 Accepted Submission(s): 291

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

题目分析 : 一道二维树状数组 的裸题, 只是需要对坐标做些处理即可, 另外, 初始化的时候 原来 com[i][j] = lowbit (i) * lowbit (j);

WA 好多次, 直接用的modify(i,j,1)

好了, 直接代码吧, 代码过长, 内存多了一点点 , HDU 第二

2

HUT-MiYu

156MS

8044K

3172B

C++

/*

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋
http://www.cnblog.com/MiYu
Author By : MiYu

Test : 1

Program : 1892

*/

#include <iostream>

#include <cmath>

#include <algorithm>

using namespace std;

#define lowbit(x) (x&(-x))

int T,N;

const int MAX = 1001;

int mat[1002][1002];

int com[1002][1002];

void modify ( int x,int y, int n )

{

while ( x <= MAX ){

int t = y;

while ( t <= MAX ){

com[x][t] += n;

t += lowbit(t);

}

x += lowbit(x);

}

}

int quy ( int x, int y )

{

int sum = 0;

while ( x > 0 ){

int t = y;

while ( t > 0 ){

sum += com[x][t];

t -= lowbit(t);

}

x -= lowbit(x);

}

return sum;

}

inline bool scan_d(int &num)

{

char in;bool IsN=false;

in=getchar();

if(in==EOF) return false;

while(in!='-'&&(in<'0'||in>'9')) in=getchar();

if(in=='-'){ IsN=true;num=0;}

else num=in-'0';

while(in=getchar(),in>='0'&&in<='9'){

num*=10,num+=in-'0';

}

if(IsN) num=-num;

return true;

}

int main ()

{

scan_d(T);{

int ca = 1;

while ( T -- ){

printf ( "Case %d:\n",ca++ );

scan_d(N); char s[5]; int a,b,x,y,m,res,maxx,maxy,minx,miny;

for ( int i = 1; i <= MAX; ++ i )

for ( int j = 1; j <= MAX; ++ j )

com[i][j] = lowbit(i) * lowbit(j), mat[i][j] = 1;

for ( int i = 1; i <= N; ++ i ){

scanf ( "%s",s );

switch ( s[0] ){

case 'S' : scan_d(a);scan_d(b);scan_d(x);scan_d(y); minx = min ( a,x );miny=min(b,y);maxx=max(a,x)+1;maxy=max(b,y)+1;

res = 0; res += quy( maxx,maxy ); res -= quy (maxx,miny); res -= quy(minx,maxy); res += quy(minx,miny);

printf ( "%d\n",res ); break;

case 'A' : scan_d(x);scan_d(y);scan_d(a);x++;y++; modify ( x,y,a ); mat[x][y] += a; break;

case 'D' : scan_d(x);scan_d(y);scan_d(a);x++;y++; if ( mat[x][y] >= a ) { modify ( x,y,-a ); mat[x][y] -= a; }

else { modify ( x,y,-mat[x][y] ); mat[x][y] = 0; } break;

case 'M' : scan_d(a);scan_d(b);scan_d(x);scan_d(y);scan_d(m);a++;b++;x++;y++; if ( mat[a][b] >= m )

{ mat[a][b] -= m; mat[x][y] += m; modify ( a,b,-m ); modify ( x,y,m ); }

else { modify ( a,b,-mat[a][b] ); modify ( x,y,mat[a][b] ); mat[x][y] += mat[a][b]; mat[a][b] = 0; } break;

}

}

}

}

return 0;

}