POJ 1068 Parencodings

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23163		Accepted: 13574

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

题目大意

给你一个数组P，pi 等于 第i个右括号前面有多少个左括号

要你求出对应的W数组，wi 等于 第i个右括号与其对应的左括号中有多少个左括号

思路： 模拟，之后根据模拟所得，计算并且输出对应的W数组

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
using namespace std;
#define inf 0xffffff
#define maxn 22

int dis[maxn],ans[maxn],vis[maxn*2],n;
//dis储存输入数据，ans储存输出数据，vis储存模拟

void fun()
{
int flag = 0,top = 0;

//进行模拟
for(int i = 0; i < n; i++){
for(int j = flag;j < dis[i]; j++) {vis[top++] = -1;}
flag = dis[i];
vis[top++] = 1;
}

/*****************************
for(int i = 0; i < top; i++)
cout<<(vis[i]>0? ')':'(');
cout<<endl;
*****************************/

flag = 0;
for(int i = 0; i < top; i++){
if(vis[i] == 1){
int dix = 1,j = i - 1,x = 1;
while(x){
x += vis[j--];
dix++;
}
ans[flag++] = dix/2;
}
}
}

int main()
{
int t;
scanf("%d",&t);
while(t--){
cin>>n;
for(int i = 0; i < n; i++) cin>>dis[i];
fun();
//这里是个坑，要比平时多打一个空格
for(int i = 0; i < n; i++) cout<<ans[i]<<' ';
cout<<endl;
}
return 0;
}