Parencodings
Time Limit: 1000 MS Memory Limit: 10000 KB
64-bit integer IO format: %I64d , %I64u Java class name: Main
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
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.
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
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stack>
using namespace std;
int main()
{
int t,n,temp;
int a[1000];
int val[1000],w[1000];
cin>>t;
while(t--)
{
memset(val,0,sizeof(val));
cin>>n;
for(int j=0; j<n; j++)
{
cin>>a[j];
temp=a[j]+j;
val[temp]=1; ///让）==1 （==0
/*for(int i=0;i<n*2;i++) ///输出1的地方
cout<<val[i]<<‘ ‘;
cout<<endl;*/
///cout<<val[temp]<<endl; ==1 从后往前数）的个数
for(int i=temp-1,sum=1,falg=0; i>=0; i--) /// val[i]==) flag标记（是否与）匹配的
{
if(falg==0&&val[i]==0) ///eg:)
{
w[j]=sum;
break;
}
else if(val[i]==0) ///eg:((())
{
falg--;
}
else if(val[i]==1) ///eg:)))
{
sum++;
falg++;
}
}
}
for(int i=0; i<n; i++)
{
cout<<w[i]<<‘ ‘;
}
cout<<endl;
}
return 0;
}
注释：
题意：P:从左到右 遇见第i个完整括号开始 （的个数 q：从左到右遇见第一个）时 与该）匹配的括号内 ）的个数 （包含该））
1： 因为不用输出括号 所以将（标记为0 ）标记为1
2：对于q 要计算一个大括号内的全部） 所以要从后向前遍历
poj1068 模拟