POJ 3225 Help with Intervals
2012-07-19 14:25
375 查看
Help with Intervals
Description
LogLoader, Inc. is a company specialized in providing products for analyzing logs. While Ikki is working on graduation design, he is also engaged in an internship at LogLoader. Among his tasks, one is to write a module for manipulating time intervals, which have confused him a lot. Now he badly needs your help.
In discrete mathematics, you have studied several basic set operations, namely union, intersection, relative complementation and symmetric difference, which naturally apply to the specialization of sets as intervals.. For your quick reference they are summarized in the table below:
Ikki has abstracted the interval operations emerging from his job as a tiny programming language. He wants you to implement an interpreter for him. The language maintains a set S, which starts out empty and is modified as specified by the following commands:
Input
The input contains exactly one test case, which consists of between 0 and 65,535 (inclusive) commands of the language. Each command occupies a single line and appears like
where
End of file (EOF) indicates the end of input.
Output
Output the set S as it is after the last command is executed as the union of a minimal collection of disjoint intervals. The intervals should be printed on one line separated by single spaces and appear in increasing order of their endpoints. If S is empty, just print “
Sample Input
Sample Output
Source
PKU Local 2007 (POJ Monthly--2007.04.28), frkstyc
//第一次遇到线段树这么用、偶数点代表点,奇数代表两点之间的区间,所以每个点要乘以2
//还有成段更新中的 up,把子节点信息传个父节点,down,把父节点的信息传递给子节点
//还有根据节点的信息作出相应的操作
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
#define N 131073
using namespace std;
short cover[N<<2];//代表该段都被覆盖成0或1
short Xor[N<<2],flag;//异或操作,代表从这起于下节点做异或操作
bool hash
;
void axor(int k)
{
if(cover[k]!=-1)
cover[k]^=1;
else
Xor[k]^=1;
}
void down(int &k)
{
if(cover[k]!=-1)
{
cover[k<<1]=cover[k<<1|1]=cover[k];
Xor[k]=0;
cover[k]=-1;
}
else if(Xor[k])
{
axor(k<<1);
axor(k<<1|1);
Xor[k]=0;
}
}
void Cover(int L,int R,int l,int r,int k)
{
if(L<=l&&R>=r)
{
cover[k]=flag;
return ;
}
down(k);
int m=(l+r)>>1;
if(L<=m) Cover(L,R,lson);
if(R>m) Cover(L,R,rson);
}
void XOr(int &L,int &R,int l,int r,int k)
{
if(L<=l&&R>=r)
{
if(cover[k]!=-1)
cover[k]^=1;
else
Xor[k]^=1;
return ;
}
down(k);
int m=(l+r)>>1;
if(L<=m) XOr(L,R,lson);
if(R>m) XOr(L,R,rson);
}
void query(int l,int r,int k)
{
if(cover[k]==1)
{ flag=0;
for(int i=l;i<=r;i++)
hash[i]=true;
return ;
}
if(cover[k]==0)
return ;
down(k);
int m=(l+r)>>1;
query(lson);
query(rson);
}
int main()
{
int a,b;
// freopen("in.txt","r",stdin);//开始这个没有注释掉、POJ不报编译错误,报WA,唉、郁闷,WA了好多次
char op,l,r;
while(scanf("%c %c%d,%d%c",&op,&l,&a,&b,&r)!=EOF)
{ getchar();
a=a<<1;b=b<<1;
if(l=='(') a++;
if(r==')') b--;
if(a>b)
{ if(op=='I'||op=='C')
cover[1]=Xor[1]=0;
continue;
}
switch(op)
{
case 'U':flag=1;Cover(a,b,0,N,1);break;
case 'I':flag=0;if(a>0) Cover(0,a-1,0,N,1);
Cover(b+1,N,0,N,1);break;
case 'D':flag=0;Cover(a,b,0,N,1);break;
case 'C':flag=0;
if(a>0) Cover(0,a-1,0,N,1);
Cover(b+1,N,0,N,1);XOr(a,b,0,N,1); break;
case 'S':XOr(a,b,0,N,1); break;
}
}
flag=1;
query(0,N,1);
int s=-1,e;
bool f=0;
if(flag)
printf("empty set\n");
else
{
for(int i=0;i<=N;i++)
if(hash[i])
{
if(s==-1)
s=i;
}
else
{
if(s!=-1)
{ e=i-1;
if(f) printf(" ");
f=1;
printf("%c%d,%d%c",s&1?'(':'[',s>>1,(e+1)>>1,e&1?')':']');
s=-1;
}
}
printf("\n");
}
return 0;
}
Time Limit: 6000MS | Memory Limit: 131072K | |
Total Submissions: 6278 | Accepted: 1391 | |
Case Time Limit: 2000MS |
LogLoader, Inc. is a company specialized in providing products for analyzing logs. While Ikki is working on graduation design, he is also engaged in an internship at LogLoader. Among his tasks, one is to write a module for manipulating time intervals, which have confused him a lot. Now he badly needs your help.
In discrete mathematics, you have studied several basic set operations, namely union, intersection, relative complementation and symmetric difference, which naturally apply to the specialization of sets as intervals.. For your quick reference they are summarized in the table below:
Operation | Notation | Definition |
---|---|---|
Union | A ∪ B | {x : x ∈ A or x ∈ B} |
Intersection | A ∩ B | {x : x ∈ A and x ∈ B} |
Relative complementation | A − B | {x : x ∈ A but x ∉ B} |
Symmetric difference | A ⊕ B | (A − B) ∪ (B − A) |
Ikki has abstracted the interval operations emerging from his job as a tiny programming language. He wants you to implement an interpreter for him. The language maintains a set S, which starts out empty and is modified as specified by the following commands:
Command | Semantics |
---|---|
UT | S ← S ∪ T |
IT | S ← S ∩ T |
DT | S ← S − T |
CT | S ← T − S |
ST | S ← S ⊕ T |
Input
The input contains exactly one test case, which consists of between 0 and 65,535 (inclusive) commands of the language. Each command occupies a single line and appears like
XT
where
Xis one of ‘
U’, ‘
I’, ‘
D’, ‘
C’ and ‘
S’ and T is an interval in one of the forms
(a
,b
),
(a
,b
],
[a
,b
)and
[a
,b
](a, b ∈ Z, 0 ≤ a ≤ b ≤ 65,535), which take their usual meanings. The commands are executed in the order they appear in the input.
End of file (EOF) indicates the end of input.
Output
Output the set S as it is after the last command is executed as the union of a minimal collection of disjoint intervals. The intervals should be printed on one line separated by single spaces and appear in increasing order of their endpoints. If S is empty, just print “
empty set” and nothing else.
Sample Input
U [1,5] D [3,3] S [2,4] C (1,5) I (2,3]
Sample Output
(2,3)
Source
PKU Local 2007 (POJ Monthly--2007.04.28), frkstyc
//第一次遇到线段树这么用、偶数点代表点,奇数代表两点之间的区间,所以每个点要乘以2
//还有成段更新中的 up,把子节点信息传个父节点,down,把父节点的信息传递给子节点
//还有根据节点的信息作出相应的操作
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define lson l,m,k<<1
#define rson m+1,r,k<<1|1
#define N 131073
using namespace std;
short cover[N<<2];//代表该段都被覆盖成0或1
short Xor[N<<2],flag;//异或操作,代表从这起于下节点做异或操作
bool hash
;
void axor(int k)
{
if(cover[k]!=-1)
cover[k]^=1;
else
Xor[k]^=1;
}
void down(int &k)
{
if(cover[k]!=-1)
{
cover[k<<1]=cover[k<<1|1]=cover[k];
Xor[k]=0;
cover[k]=-1;
}
else if(Xor[k])
{
axor(k<<1);
axor(k<<1|1);
Xor[k]=0;
}
}
void Cover(int L,int R,int l,int r,int k)
{
if(L<=l&&R>=r)
{
cover[k]=flag;
return ;
}
down(k);
int m=(l+r)>>1;
if(L<=m) Cover(L,R,lson);
if(R>m) Cover(L,R,rson);
}
void XOr(int &L,int &R,int l,int r,int k)
{
if(L<=l&&R>=r)
{
if(cover[k]!=-1)
cover[k]^=1;
else
Xor[k]^=1;
return ;
}
down(k);
int m=(l+r)>>1;
if(L<=m) XOr(L,R,lson);
if(R>m) XOr(L,R,rson);
}
void query(int l,int r,int k)
{
if(cover[k]==1)
{ flag=0;
for(int i=l;i<=r;i++)
hash[i]=true;
return ;
}
if(cover[k]==0)
return ;
down(k);
int m=(l+r)>>1;
query(lson);
query(rson);
}
int main()
{
int a,b;
// freopen("in.txt","r",stdin);//开始这个没有注释掉、POJ不报编译错误,报WA,唉、郁闷,WA了好多次
char op,l,r;
while(scanf("%c %c%d,%d%c",&op,&l,&a,&b,&r)!=EOF)
{ getchar();
a=a<<1;b=b<<1;
if(l=='(') a++;
if(r==')') b--;
if(a>b)
{ if(op=='I'||op=='C')
cover[1]=Xor[1]=0;
continue;
}
switch(op)
{
case 'U':flag=1;Cover(a,b,0,N,1);break;
case 'I':flag=0;if(a>0) Cover(0,a-1,0,N,1);
Cover(b+1,N,0,N,1);break;
case 'D':flag=0;Cover(a,b,0,N,1);break;
case 'C':flag=0;
if(a>0) Cover(0,a-1,0,N,1);
Cover(b+1,N,0,N,1);XOr(a,b,0,N,1); break;
case 'S':XOr(a,b,0,N,1); break;
}
}
flag=1;
query(0,N,1);
int s=-1,e;
bool f=0;
if(flag)
printf("empty set\n");
else
{
for(int i=0;i<=N;i++)
if(hash[i])
{
if(s==-1)
s=i;
}
else
{
if(s!=-1)
{ e=i-1;
if(f) printf(" ");
f=1;
printf("%c%d,%d%c",s&1?'(':'[',s>>1,(e+1)>>1,e&1?')':']');
s=-1;
}
}
printf("\n");
}
return 0;
}
相关文章推荐
- poj 3225 线段树区间更新,区间询问 Help with Intervals
- POJ 3225 Help with Intervals(线段树成段更新)
- poj 3225 Help with Intervals(线段树进阶,处理区间,拆点)
- POJ 3225 - Help with Intervals
- poj 3225 Help with Intervals(线段树+区间的交集,差集,补集,并集)好难的题目,一天了,真是纠结
- POJ 3225 Help with Intervals 线段树
- POJ 3225 Help with Intervals(区间更新 + 倍增区间)
- poj 3225 Help with Intervals(线段树)
- POJ 3225 Help with Intervals
- POJ 3225 Help with Intervals(线段树)
- POJ 3225 Help with Intervals
- poj 3225 Help with Intervals
- 【POJ】3225 Help with Intervals
- POJ 3225 Help with Intervals(线段树)
- poj 3225 Help with Intervals(线段树)
- poj 3225 Help with Intervals(线段树区间更新)
- POJ 3225 Help with Intervals(开区间和闭区间)
- poj 3225 Help with Intervals -线段树-延迟标记-区间交并补
- POJ 3225 - Help with Intervals
- poj 3225 Help with Intervals