POJ-3225 Help with Intervals

Help with Intervals

Time Limit: 6000MS Memory Limit: 131072KCase 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:

OperationNotation

Definition

UnionA ∪ B{x : x ∈ A or x ∈ B}IntersectionA ∩ B{x : x ∈ A and x ∈ B}Relative complementationA − B{x : x ∈ A but x ∉ B}Symmetric differenceA ⊕ 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 setS, which starts out empty and is modified as specified by the following commands:

CommandSemanticsU TS ← S ∪ TI TS ← S ∩ TD TS ← S − TC TS ← T − SS TS ← 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

X T

where X is one of ‘U’, ‘I’, ‘D’, ‘C’ and ‘S’ andT 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. IfS 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)

————————————————————集训26.1的分割线————————————————————

前言：补考的准备算是差不多了，毕竟放弃了傅里叶级数和热力学。。。根本看不懂。。。回到实验室。

思路：区间交并补等的线段树模板吧……根据胡大大的理解

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

重点是如何处理开区间和闭区间。

线段树是用点来表示区间的一种数据结构。

要知道像是(2, 3)这种区间并不是空区间。要表示开区间和闭区间，只需要把每个点都拆成两个点——拆点，然后(2, 3)就变成了[5, 5]，[2, 3]就变成了[4, 6]。如下：

[a (a , b) b]

[a*2 (a*2+1 , b*2-1) b*2]

类比一下哈希的思想就是：

2    2 ~ 3    3

↓       ↓      ↓

4       5       6

代码如下：

/*ID: j.sure.1PROG:LANG: C++*//****************************************/#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <stack>#include <queue>#include <vector>#include <map>#include <string>#include <climits>#include <iostream>#define INF 0x3f3f3f3fusing namespace std;/****************************************/#define lson l, m, rt<<1#define rson m+1, r, rt<<1|1const int maxn = 2e5;//65535*2+2;int cover[maxn<<2];bool oppo[maxn<<2], vis[maxn];void toOP(int rt){//翻转操作if(cover[rt] != -1) {cover[rt] = !cover[rt];}else oppo[rt] = !oppo[rt];}void Down(int rt){if(cover[rt] != -1) {cover[rt<<1] = cover[rt<<1|1] = cover[rt];oppo[rt<<1] = oppo[rt<<1|1] = 0;cover[rt] = -1;}//没有覆盖标记但是有翻转标记if(oppo[rt]) {toOP(rt<<1); toOP(rt<<1|1);oppo[rt] = 0;}}void update(char op, int L, int R, int l, int r, int rt){if(L <= l && r <= R) {switch(op) {case 'U': cover[rt] = 1; oppo[rt] = 0; break;case 'D': cover[rt] = 0; oppo[rt] = 0; break;case 'S': toOP(rt); break;//翻转01case 'C': toOP(rt); break;//翻转01而且补集用0覆盖}return ;}Down(rt);int m = (l+r)>>1;if(L <= m) update(op, L, R, lson);else {//否则就是补集if(op == 'I' || op == 'C') {cover[rt<<1] = oppo[rt<<1] = 0;}//覆盖补集}if(m < R) update(op, L, R, rson);else {if(op == 'I' || op == 'C') {cover[rt<<1|1] = oppo[rt<<1|1] = 0;}}}void Cnt(int l, int r){for(int i = l; i <= r; i++) {vis[i] = true;}}void query(int l, int r, int rt){if(cover[rt] == 1) {Cnt(l, r);//数区间return ;} else if(!cover[rt]) return ;//就像是剪枝一样if(l == r) return;//边界Down(rt);int m = (l+r) >> 1;query(lson); query(rson);}int main(){#ifdef J_Sure//freopen("000.in", "r", stdin);//freopen(".out", "w", stdout);#endifchar op, ll, rr;int a, b;while(~scanf("%c", &op)) {#ifdef J_Sureif(op == 'E') break;#endifscanf(" %c%d,%d%c\n", &ll, &a, &b, &rr);a <<= 1; b <<= 1;if(ll == '(') a ++;if(rr == ')') b --;//处理开闭区间的方法：拆点update(op, a, b, 0, maxn, 1);}query(0, maxn, 1);bool ept = true;char c;//接下来是数区间for(int i = 0; i <= maxn; ) {if(vis[i]) {c = i&1 ? '(' : '[';if(!ept) printf(" ");ept = false;printf("%c%d,", c, i/2);while(vis[i+1]) i++;c = i&1 ? ')' : ']';printf("%d%c", (i+1)/2, c);}i++;}if(ept) puts("empty set");return 0;}