HDOJ 2642 HDU 2642 Stars ACM 2642 IN HDU
4984 ワード
MiYuオリジナル、転帖は明記してください:転載は______________白い家 
から
タイトルアドレス:
http://acm.hdu.edu.cn/showproblem.php?pid=2642
タイトルの説明:
StarsTime Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)Total Submission(s): 166 Accepted Submission(s): 66
Problem Description
Yifenfei is a romantic guy and he likes to count the stars in the sky.
To make the problem easier,we considerate the sky is a two-dimension plane.Sometimes the star will be bright and sometimes the star will be dim.At first,there is no bright star in the sky,then some information will be given as "B x y"where 'B' represent bright and x represent the X coordinate and y represent the Y coordinate means the star at (x,y) is bright,And the 'D' in "D x y"mean the star at(x,y) is dim.When get a query as "Q X1 X2 Y1 Y2",you should tell Yifenfei how many bright stars there are in the region correspond X1,X2,Y1,Y2.
There is only one case.
Input
The first line contain a M(M <= 100000), then M line followed.
each line start with a operational character.
if the character is B or D,then two integer X,Y (0 <=X,Y<= 1000)followed.
if the character is Q then four integer X1,X2,Y1,Y2(0 <=X1,X2,Y1,Y2<= 1000) followed.
Output
For each query,output the number of bright stars in one line.
Sample Input
Sample Output
タイトル分析:HDU 1892と基本的に同じ問題で、マトリクス配列の値が[0,1]に限定されているだけで、2次元樹状配列の裸問題であり、直接テンプレートであり、テーマデータに対していくつかの処理を加えればよい.コードは次のとおりです.
/*
MiYuオリジナル、転帖は明記してください:転載は_______白い家
http://www.cnblog.com/MiYu
Author By : MiYu
Test : 1
Program : 2642
*/
#include
#include
using namespace std;
#define lowbit(x) (x&(-x))
int T;
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 ()
{
while ( scan_d(T) ) {
int ca = 1;
char s[5]; int a,b,x,y,m,res,maxx,maxy,minx,miny;
memset ( com, 0, sizeof ( com ) );
while ( T -- ) {
scanf ( "%s",s );
switch ( s[0] ){
case 'Q' : scan_d(a);scan_d(x);scan_d(b);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",res ); break;
case 'B' : scan_d(x);scan_d(y); x++;y++; if ( !mat[x][y] ) { modify ( x,y,1 ); mat[x][y] = 1; } break;
case 'D' : scan_d(x);scan_d(y); x++;y++; if ( mat[x][y] ) { modify ( x,y,-1 ); mat[x][y] = 0; } break;
}
}
}
return 0;
}
からタイトルアドレス:
http://acm.hdu.edu.cn/showproblem.php?pid=2642
タイトルの説明:
StarsTime Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/65536 K (Java/Others)Total Submission(s): 166 Accepted Submission(s): 66
Problem Description
Yifenfei is a romantic guy and he likes to count the stars in the sky.
To make the problem easier,we considerate the sky is a two-dimension plane.Sometimes the star will be bright and sometimes the star will be dim.At first,there is no bright star in the sky,then some information will be given as "B x y"where 'B' represent bright and x represent the X coordinate and y represent the Y coordinate means the star at (x,y) is bright,And the 'D' in "D x y"mean the star at(x,y) is dim.When get a query as "Q X1 X2 Y1 Y2",you should tell Yifenfei how many bright stars there are in the region correspond X1,X2,Y1,Y2.
There is only one case.
Input
The first line contain a M(M <= 100000), then M line followed.
each line start with a operational character.
if the character is B or D,then two integer X,Y (0 <=X,Y<= 1000)followed.
if the character is Q then four integer X1,X2,Y1,Y2(0 <=X1,X2,Y1,Y2<= 1000) followed.
Output
For each query,output the number of bright stars in one line.
Sample Input
5
B 581 145
B 581 145
Q 0 600 0 200
D 581 145
Q 0 600 0 200
Sample Output
1
0
タイトル分析:HDU 1892と基本的に同じ問題で、マトリクス配列の値が[0,1]に限定されているだけで、2次元樹状配列の裸問題であり、直接テンプレートであり、テーマデータに対していくつかの処理を加えればよい.コードは次のとおりです.
/*
MiYuオリジナル、転帖は明記してください:転載は_______白い家
http://www.cnblog.com/MiYu
Author By : MiYu
Test : 1
Program : 2642
*/
#include
#include
using namespace std;
#define lowbit(x) (x&(-x))
int T;
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 ()
{
while ( scan_d(T) ) {
int ca = 1;
char s[5]; int a,b,x,y,m,res,maxx,maxy,minx,miny;
memset ( com, 0, sizeof ( com ) );
while ( T -- ) {
scanf ( "%s",s );
switch ( s[0] ){
case 'Q' : scan_d(a);scan_d(x);scan_d(b);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",res ); break;
case 'B' : scan_d(x);scan_d(y); x++;y++; if ( !mat[x][y] ) { modify ( x,y,1 ); mat[x][y] = 1; } break;
case 'D' : scan_d(x);scan_d(y); x++;y++; if ( mat[x][y] ) { modify ( x,y,-1 ); mat[x][y] = 0; } break;
}
}
}
return 0;
}