Gym - 101190H------Hard Refactoring（字符串 模拟）
http://codeforces.com/gym/101190/attachments
Input file:
hard.in
Output file:
hard.out
Helen had come upon a piece of code that uses a lot of “magical constants”. She found a logical expression that checks if an integer x belongs to a certain set of ranges, like the one shown below:
x >= 5 && x <= 10 ||
x >= 7 && x <= 20 ||
x <= 2 ||
x >= 21 && x <= 25 ||
x >= 8 && x <= 10 ||
x >= 100
Helen does not like “magical constants”, so she decided to refactor this expression and all similar ones in such a way, that the refactored expression still has the same Boolean result for all integers x, but it uses as few integer constants in its text as possible.
Integers in this problem, including integer x, come from the range of all signed 16 bit integers starting from −2 15 (−32 768) to 215 − 1 (32 767) inclusive.
Input
The input file contains at most 1000 lines. Each line consists of either one comparison or two comparisons separated by logical and operator “&&”. Each comparison starts with “x”, followed by greater-or-equals operator “>=” or less-or-equals operator “<=”, followed by an integer constant. When two comparisons are in the same line, the first one is always greater-or-equals, followed by less-or-equals.
All lines, but the last one, are terminated by logical or operator “||”. All tokens in a line are separated by a single space and there are no trailing or leading spaces.
Output
Write the refactored expression to the output file in the same format as in the input. You can arrange lines in any order, as long as the resulting expression has the right format, produces the same Boolean result on all integers x, and contains the minimal possible number of integer constants in its text. Numbers must be formatted without leading zeros and there must be precisely one space between tokens on a line.
Write a single line with the word “true” if the expression is true on all integers. Write a single line with the word “false” if the expression is false on all integers.
Examples
hard.in
x >= 5 && x <= 10 ||
x >= 7 && x <= 20 ||
x <= 2 ||
x >= 21 && x <= 25 ||
x >= 8 && x <= 10 ||
x >= 100
hard.out
x <= 2 ||
x >= 5 && x <= 25 ||
x >= 100
hard.in
x >= 10 && x <= 0
hard.out
false
hard.in
x <= 10 ||
x >= 0
hard.out
true
hard.in
x >= -32768
hard.out
true
题意：
[−32768,32767]这是总区间，
求的是所给区间在[−32768,32767]上的区间，
没有输出false，总区间包含于所给区间输出true，否则输出所求区间
思路：
排下序，能并起来的并起来，并不起来的就当是新区间，
坑不少，比赛的时候WA了好久也没写对，
这里我加几个样例填下坑：
input:
x >= 0 && x <= 10 ||
x >= 90
output:
x >= 90
input:
x <= -32768
output:
x <= -32768
input:
x <= -32767
output:
false
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<algorithm> 5 #include<iostream> 6 7 const int init_left = -32768; 8 const int init_right = 32767; 9 10 using namespace std; 11 12 struct node 13 { 14 int left, right; 15 } str[50000], init; 16 17 int judge(char c) 18 { 19 if(c>='0'&&c<='9') return 1; 20 else return 0; 21 } 22 23 int cmp(struct node a, struct node b) 24 { 25 if(a.left != b.left) return a.left < b.left; 26 else return a.right > b.right; 27 } 28 29 30 int main() 31 { 32 freopen("hard.in","r",stdin); 33 freopen("hard.out","w",stdout); 34 int len, i, num, j, x; 35 char a[10005]; 36 num = 0; 37 init.left = init_left; 38 init.right = init_right; 39 while(1) 40 { 41 str[num] = init; 42 gets(a); 43 len = strlen(a); 44 for(i=0; i<len; i++) 45 { 46 if(a[i]=='=') 47 { 48 for(j=i; j<len; j++) 49 { 50 if(judge(a[j])) 51 { 52 x = 0; 53 while(judge(a[j])) 54 { 55 x = x * 10 + a[j]-'0'; 56 j++; 57 } 58 if(a[i+1]=='-'||a[i+2]=='-'||a[i+3]=='-') x = -x; 59 if(a[i-1]=='<') str[num].right = x; 60 else str[num].left = x; 61 break; 62 } 63 } 64 } 65 } 66 num++; 67 if(a[len-1]!='|') break; 68 } 69 sort(str, str+num, cmp); 70 int cur = 0, sum = 0; 71 for(i=0; i<num; i++) 72 { 73 if(str[i].left>str[i].right) 74 { 75 sum++; 76 continue; 77 } 78 if(str[i].left<=str[cur].right+1) str[cur].right = max(str[cur].right, str[i].right); 79 else 80 { 81 cur++; 82 str[cur] = str[i]; 83 } 84 } 85 86 if(cur==0&&str[0].left==init_left&&str[0].right==init_right) printf("true\n"); 87 else if(sum==num) printf("false\n"); 88 else 89 { 90 for(i=0; i<=cur; i++) 91 { 92 if(str[i].left<=init_left) printf("x <= %d", str[i].right); 93 else if(str[i].right>=init_right) printf("x >= %d", str[i].left); 94 else printf("x >= %d && x <= %d", str[i].left, str[i].right); 95 if(i==cur) printf("\n"); 96 else printf(" ||\n"); 97 } 98 } 99 return 0; 100 }
