POJ 1068 Parencodings【水模拟--数括号】

链接：

http://poj.org/problem?id=1068

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27454#problem/B

Parencodings

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 17044		Accepted: 10199

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

Source

Tehran 2001

题意：

先看题目样例中给出的括号序列　S　注意到它有 6 个右括号

而下面的两个数字序列 Ｐ 和 Ｗ 都含有 ６ 个数字，而且都写在这右边６个括号下面

再仔细看序列 P 代表的是从左到右,每一个右括号前面的左括号的个数

平时生活中括号都是一对对存在的

W序列代表的是与当前右括号匹配的最近的左括号编号

【这个编号是按照离开当前需要匹配的右边括号的,左边括号个数计算的】

题目给了你序列 P 要你输出序列 W ,简单模拟一下就好了

算法：

简单模拟

思路：

用一个数组标记即可,左括号标记为 0,右括号标记为 1

同时再用一个 vis[] 数组标记已经被匹配过了的左边括号

/**
Accepted	180K	0MS	C++	1586B
题意：先看题目样例中给出的括号序列　S　注意到它有 6 个右括号
而下面的两个数字序列 Ｐ 和 Ｗ 都含有 ６ 个数字，而且都写在这右边６个括号下面
再仔细看序列 P 代表的是从左到右,每一个右括号前面的左括号的个数
平时生活中括号都是一对对存在的
W序列代表的是与当前右括号匹配的最近的左括号编号
【这个编号是按照离开当前需要匹配的右边括号的,左边括号个数计算的】

题目给了你序列 P 要你输出序列 W ,简单模拟一下就好了

算法：简单模拟

思路：用一个数组标记即可,左括号标记为 0,右括号标记为 1
同时再用一个 vis[] 数组标记已经被匹配过了的左边括号

*/
#include<stdio.h>
#include<string.h>
const int maxn = 2000;

int a[maxn]; //记录括号是左边还是右边
int vis[maxn]; // 标记是否被匹配

int main()
{
int T;
int n;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
memset(vis, 0, sizeof(vis)); //初始化清空
memset(a, 0, sizeof(a));
int x;
int t = 0; //记录当下左括号数目
int index = 1; //记录当前括号下标
int j;
for(int i = 0; i < n; i++)
{
scanf("%d", &x); //每次输入一个右边括号,同时补全它左边括号的数目
if(x > t)
{
int tmp = x-t; //记录当前左边
t = x;
while(tmp--)
{
a[index++] = 0; //补全左边括号
}
}
a[index++] = 1; //记录当前右边括号
}

int flag = 0;
int num = index; // 总的括号数目
//for(int i = 1; i < num; i++) printf("%d ", a[i]); printf("\n");
for(int i = 1; i < num; i++) // 从前往后遍历 ）
{
if(a[i] == 1) // )
{
flag = 0;
int ans = 0;
for(int j = i-1; j >= 1; j--) // 从后往前匹配
{
if(flag == 1) break;
if(a[j] == 0 && vis[j] && flag == 0)
{
ans++;
}
else if(a[j] == 0 && !vis[j])
{
ans++;
vis[j] = 1;
printf("%d ", ans);
flag = 1;
break;
}
}
}
}
printf("\n");
}
return 0;
}