Help with Intervals
| Time Limit: 6000MS | Memory Limit: 131072K | |
| Total Submissions: 12474 | Accepted: 3140 | |
| Case Time Limit: 2000MS | ||
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:
Operation Notation
DefinitionUnion 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 UTS ← S ∪ T ITS ← S ∩ T DTS ← S ? T CTS ← T ? S STS ← 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 X is 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
很久没做线段树了，拿来练手一脸懵逼。
这题坑点太多了，空集情况要考虑（这个坑了好久），边界要考虑（你以为是真无穷吗那不是药丸）
题解摘自kungbin博客：http://www.cnblogs.com/kuangbin/archive/2013/04/10/3012986.html
题意：给一个全局为0~65536的区间，一开始区间s为空间，然后不断地对区间s进行并上一个区间，交一个区间，减一个区间，用一个区间减去s，还有异或下两个区间。。。
题意:区间操作,交,并,补等
思路:
我们一个一个操作来分析:(用0和1表示是否包含区间,-1表示该区间内既有包含又有不包含)
U:把区间[l,r]覆盖成1
I:把[-∞,l)(r,∞]覆盖成0
D:把区间[l,r]覆盖成0
C:把[-∞,l)(r,∞]覆盖成0 , 且[l,r]区间0/1互换
S:[l,r]区间0/1互换
分析：这题的各个操作都可以用线段树的成段更新在log（65536*2）的时间内完成
U：并上一个区间，也就是，将这个区间置1就行
I：交上一个区间[l,r]，将区间[-∞,l)和(r,∞]置0
D：减去一个区间，将这个区间置0就行
C：用一个区间[l,r]减去s，将区间[-∞,l)和(r,∞]置0，区间[l,r]取反就行
S：求异或，区间[l,r]取反就行
现在上面的所有操作都在理论上解决掉了，而区间的开或闭这个直接把所有区间乘2，对于左边的开区间要加1，右边减1。
记得！空集！还有边界的0!MAXN!
1 #include<cstdio>
2 #include<iostream>
3 #include<cstring>
4 #define clr(x) memset(x,0,sizeof(x))
5 const int MAXN=65536*2;
6 using namespace std;
7 struct segtree
8 {
9 int l,r,rev,val;// val==0表示全部没有覆盖，val==1表示全部被覆盖，其余为-1,rev==1表示取反,rev==0表示不变，只有val==-1时候rev才有作用
10 }tree[MAXN*4+10];
11 int cov[MAXN*2+10];
12 void init(int l,int r,int i)//初始化
13 {
14 tree[i].l=l;
15 tree[i].r=r;
16 tree[i].rev=tree[i].val=0;
17 if(l==r) return ;
18 int mid=(l+r)>>1;
19 init(l,mid,i<<1);
20 init(mid+1,r,(i<<1)+1);
21 return ;
22 }
23 void reverse(int i)//取反
24 {
25 if(tree[i].val!=-1) tree[i].val^=1;
26 else
27 tree[i].rev^=1;
28 }
29 void downto(int i)//向下更新
30 {
31 if(tree[i].val!=-1)
32 {
33 tree[i<<1].val=tree[(i<<1)+1].val=tree[i].val;
34 tree[i].val=-1;
35 tree[i<<1].rev=tree[(i<<1)+1].rev=0;
36 }
37 if(tree[i].rev)
38 {
39 reverse(i<<1);
40 reverse((i<<1)+1);
41 tree[i].rev=0;
42 }
43 return ;
44 }
45 void update(int i,int l,int r,int op)//操作更新区间
46 {
47 if(l<=tree[i].l && r>=tree[i].r)
48 {
49 if(op==0 || op==1)
50 {
51 tree[i].val=op;
52 tree[i].rev=0;
53 }
54 else
55 reverse(i);
56 return;
57 }
58 downto(i);
59 int mid=(tree[i].l+tree[i].r)>>1;
60 if(r<=mid)
61 update(i<<1,l,r,op);
62 else
63 if(l>mid)
64 update((i<<1)+1,l,r,op);
65 else
66 {
67 update((i<<1),l,r,op);
68 update((i<<1)+1,l,r,op);
69 }
70 return ;
71 }
72 void query(int i)//标识出哪些区间存在，cov数组用来记录
73 {
74 if(tree[i].val==1)
75 {
76 for(int j=tree[i].l;j<=tree[i].r;j++)
77 cov[j]=1;
78 return ;
79 }
80 if(tree[i].val==0)
81 return;
82 if(tree[i].l==tree[i].r) return ;
83 downto(i);
84 query(i<<1);
85 query((i<<1)+1);
86 return ;
87 }
88 int main()
89 {
90 int lt,rt;
91 char op,lc,rc;
92 init(0,MAXN,1);
93 while(scanf(" %c %c%d,%d%c",&op,&lc,<,&rt,&rc)!=EOF)
94 {
95 lt<<=1;
96 if(lc==‘(‘)
97 lt++;
98 rt<<=1;
99 if(rc==‘)‘)
100 rt--;
101 if(lt>rt)//空集情况！一定要注意
102 {
103 if(op==‘I‘ || op==‘C‘) update(1,0,MAXN,0);
104 }
105 else
106 {
107 if(op==‘U‘)
108 {
109 update(1,lt,rt,1);
110 }
111 if(op==‘I‘)
112 {
113 if(lt>0)update(1,0,lt-1,0);//注意边界lt>0 才能-1，rt也是如此。
114 if(rt<MAXN)update(1,rt+1,MAXN,0);
115 }
116 if(op==‘D‘)
117 {
118 update(1,lt,rt,0);
119 }
120 if(op==‘C‘)
121 {
122 if(lt>0)update(1,0,lt-1,0);
123 if(rt<MAXN)update(1,rt+1,MAXN,0);
124 update(1,lt,rt,2);
125 }
126 if(op==‘S‘)
127 {
128 update(1,lt,rt,2);
129 }
130 }
131 }
132 clr(cov);
133 query(1);
134 lt=0;
135 rt=-1;
136 for(int i=0;i<=MAXN;i++)
137 {
138 if(cov[i]==0 && cov[i+1]==1)
139 {
140 lt=i+1;
141 }
142 if(cov[i]==1 && cov[i+1]==0)
143 {
144 rt=i;
145 if(lt%2==1)
146 printf("(%d,",lt/2);
147 else
148 printf("[%d,",lt/2);
149 if(rt%2==1)
150 printf("%d) ",rt/2+1);
151 else
152 printf("%d] ",rt/2);
153 }
154 }
155 if(lt>rt)
156 {
157 printf("empty set\n");
158 }
159 else
160 printf("\n");
161 return 0;
162 }