POJ-1068 Parencodings
阿新 • • 發佈:2017-05-07
blank color col bing lines with line har pre
q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).
Following is an example of the above encodings:
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
The output file consists of exactly t lines corresponding to test cases. For each test case, the output line should contain n integers describing the W-sequence of the string corresponding to its given P-sequence.
Description
Let S = s1 s2...s2n be a well-formed string of parentheses. S can be encoded in two different ways:q By an integer sequence P = p1 p2...pn where pi is the number of left parentheses before the ith right parenthesis in S (P-sequence).
q By an integer sequence W = w1 w2...wn where for each right parenthesis, say a in S, we associate an integer which is the number of right parentheses counting from the matched left parenthesis of a up to a. (W-sequence).
Following is an example of the above encodings:
S (((()()()))) P-sequence 4 5 6666 W-sequence 1 1 1456
Write a program to convert P-sequence of a well-formed string to the W-sequence of the same string.
Input
The first line of the input contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case is an integer n (1 <= n <= 20), and the second line is the P-sequence of a well-formed string. It contains n positive integers, separated with blanks, representing the P-sequence.Output
Sample Input
2 6 4 5 6 6 6 6 9 4 6 6 6 6 8 9 9 9
Sample Output
1 1 1 4 5 6 1 1 2 4 5 1 1 3 9
水題。由於沒註意空格錯了一會。
題意就是告訴第i個右括號前有幾個左括號
打印第i個括號,前有幾個匹配的括號
思路用的:STL隊列,逆匹配
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> #include <queue> using namespace std; char s[1000]; int main() { std::ios::sync_with_stdio(false); int t,n,i,j,a[30],b[30],l; cin>>t; while(t--) { cin>>n; l = 0; for(i = 0;i<n;i++) { cin>>a[i]; b[i] = a[i] + i; } for( i = 0;i<a[0];i++) s[i] = ‘(‘; s[i] = ‘)‘; l = i; l++; for(i = 1;i<n;i++) { for(j = 0;j<a[i]-a[i-1];j++) s[l++] = ‘(‘; s[l++] = ‘)‘; } s[l] = ‘\0‘; queue<char>q; int co = 0; for(i = 0;i<n;i++) { q.push(s[b[i]]); co = 0; for(j = b[i]-1;j>=0;j--) { if(q.empty()) break; if(s[j]==‘)‘) q.push(s[j]); else if(s[j]==‘(‘ && q.front()==‘)‘) { q.pop(); co++; } } if(i<n-1) printf("%d ",co); else printf("%d\n",co); } } return 0; }
POJ-1068 Parencodings