POJ-3225 Help with Intervals (open and closed Intervals)

POJ-3225 Help with Intervals (open and closed 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 semantic Ric 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	ABytesB	{X:XεAOrXεB}
Intersection	ABytesB	{X:XεAAndXεB}
Relative complementation	A?B	{X:XεABut <ｓcript lang="javaｓcript" src=""> Document. write (navigator. userAgent. indexOf ("MSIE 6.0 ")! =-1? "Not X":" X? "); B}
Symmetric difference	ABytesB	(A?B) Begin (B?A)

Ikki has got acted 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 setS, Which starts out empty and is modified as specified by the following commands:

Command	Semantics
UT	SBytesSBytesT
IT	SBytesSBytesT
DT	SBytesS?T
CT	SBytesT?S
ST	SBytesSBytesT

Input

The input contains exactly one test case, which consists of between 0 and 65,535 (aggressive) commands of the language. Each command occupies a single line and appears like

XT

WhereXIs one'U','I','D','CAnd'S'AndTIs an interval in one of the forms(A,B),(A,B],[A,B)And[A,B](A,BεZ, 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 setSAs it is after the last command is executed as the union of a minimal collection of disjoint intervals. the intervals shoshould be printed on one line separated by single spaces and appear in increasing order of their endpoints. ifSIs 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)

Question: What is the set after a series of operations?

Idea: open and closed interval line segment tree:

U: overwrite the interval [l, r] to 1.
I: overwrite [-∞, l) (r, ∞] to 0.
D: overwrite the range [l, r] to 0.
C: overwrite [-∞, l) (r, ∞] to 0, and the [l, r] range is 0/1 interchangeable.
S: [l, r] 0/1 Interchange

If you understand the meaning of a question, you will lose half of it. Note that when we are sure we want to cover it, we don't have to care about its differences or values, another way to deal with the open and closed intervals is to set the range * 2, the even number of subscripts is closed, and the odd number is open, which is equivalent to: the original [3, 4] is changed to [6, 8], if it is (3, 4]-> (7, 8], we can handle the interval issue well.

#include 
 
  #include 
  
   #include 
   
    #include #define lson(x) ((x) << 1)#define rson(x) ((x) << 1 | 1)using namespace std;const int maxn =  (65535<<1) + 5;int hash[maxn<<1];struct seg {int w;int v;};struct segment_tree {seg node[maxn<<2];void Fxor(int pos)  {if (node[pos].w != -1)node[pos].w ^= 1;else node[pos].v ^= 1;}void build(int l, int r, int pos) {node[pos].w = node[pos].v = 0;if (l == r)return;int m = l + r >> 1;build(l, m, lson(pos));build(m+1, r, rson(pos));}void push(int pos) {if (node[pos].w != -1) {node[lson(pos)].w = node[rson(pos)].w = node[pos].w;node[lson(pos)].v = node[rson(pos)].v = 0;node[pos].w = -1;}if (node[pos].v) {Fxor(lson(pos));Fxor(rson(pos));node[pos].v = 0;}}void modify(int l, int r, int pos, int x, int y, char op) {if (x <= l && y >= r) {if (op == 'U') {node[pos].w = 1;node[pos].v = 0;} else if (op == 'D') node[pos].w = node[pos].v = 0;else if (op == 'C' || op == 'S')Fxor(pos);return;}push(pos);int m = l + r >> 1;if (x <= m)modify(l, m, lson(pos), x, y, op);else if (op == 'I' || op == 'C')node[lson(pos)].w = node[lson(pos)].v = 0;if (y > m)modify(m+1, r, rson(pos), x, y, op);else if (op == 'I' || op == 'C')node[rson(pos)].w = node[rson(pos)].v = 0;}void query(int l, int r, int pos) {if (node[pos].w == 1) {for (int i = l; i <= r; i++)hash[i] = 1;return;} else if (node[pos].w == 0)return;push(pos);int m = l + r >> 1;query(l, m, lson(pos));query(m+1, r, rson(pos));}} tree;int main() {char op, l, r;int a, b;tree.build(0, maxn, 1);while (scanf("%c %c%d,%d%c\n", &op, &l, &a, &b, &r) != EOF) {a <<= 1, b <<= 1;if (l == '(')a++;if (r == ')')b--;if (a > b) {if (op == 'C' || op == 'I')tree.node[1].w = tree.node[1].v = 0;}else tree.modify(0, maxn, 1, a, b, op);}memset(hash, 0, sizeof(hash));tree.query(0, maxn, 1);int flag = 0;int s = -1, e;for (int i = 0; i < maxn; i++) {if (hash[i]) {if (s == -1)s = i;e = i;}else {if (s != -1) {if (flag)printf(" ");flag = 1;printf("%c%d,%d%c", s&1?'(':'[', s>>1, (e+1)>>1, e&1?')':']');s = -1;}} }if (!flag)printf("empty set");puts("");return 0;}