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 9Sample Output
1 1 1 4 5 6 1 1 2 4 5 1 1 3 9Source
Tehran 2001先说说题目的意思吧，这道题是关于括号匹配的题目
首先输入一个t代表测试数据组数，紧接着输入n，代表右括号的数量
然后在输入n个数，代表每一个右括号的前面有多少个左括号
现在他的问题是，通过输入的数据判断，每一个右括号中间包括了多少对完整的括号
举个例子：(()())
第一个右括号中只括下了他自己一对括号，即1
第二个右括号和第一个情况一样，也是1
但是第三个右括号中间就扩了两个括号进去(()())
这道题，我最先开始想的是利用回溯算法来统计个数，可是怎么也没写对，杯具了很久之后想到直接统计右括号数量并且向前模拟的方法，然后很快就AC了。。后来仔细看了两份代码，发现，改的第一次写的代码中间已经用到了第二次写的一些思路，只不过大体的方向搞错了，所以一直有问题，不过从理论上讲，利用回溯法也应该可以AC，不过可能要麻烦许多。。。
AC代码：
Source Code
Problem: 1068 User: bingshenMemory: 176K Time: 0MSLanguage: C++ Result: Accepted Source Code #include<stdio.h> #include<string.h> char s[100]; int p[100]; int right[100]; int w[100]; int ans[100]; int main() { int i,n,t,l,j,num; scanf("%d",&t); while(t--) { scanf("%d",&n); memset(p,0,sizeof(p)); memset(w,0,sizeof(w)); memset(s,'(',sizeof(s)); memset(right,0,sizeof(right)); l=0; num=0; for(i=1;i<=n;i++) { scanf("%d",&p[i]); if(i==1) right[i]=p[i]+1; else right[i]=right[i-1]+(p[i]-p[i-1])+1; } for(i=1;i<=n;i++) s[right[i]]=')'; int lc=0,rc=0; for(i=1;i<=n;i++) { for(j=right[i];j>=1;j--) { if(s[j]=='(') lc++; if(s[j]==')') rc++; if(lc==rc) break; } printf("%d ",rc); lc=0; rc=0; } printf("/n"); } return 0; } 附一份用回溯写的Wrong Answer程序
Source Code
Problem: 1068 User: bingshenMemory: N/A Time: N/ALanguage: C++ Result: Wrong Answer Source Code #include<stdio.h> #include<string.h> char s[100]; int p[100]; int right[100]; int w[100]; int ans[100]; int dfs(int l,int r) { int lc=0,rc=0,i,sum=0,start; if(l==r-1) return 1; else { start=l+1; for(i=l+1;i<=r-1;i++) { if(s[i]=='(') lc++; if(s[i]==')') rc++; if(lc==rc) { w[i]=dfs(start,i); sum=sum+w[i]; lc=0; rc=0; start=i+1; } } return sum+1; } } int main() { int i,n,t,l,j,num; scanf("%d",&t); while(t--) { scanf("%d",&n); memset(p,0,sizeof(p)); memset(w,0,sizeof(w)); memset(s,'(',sizeof(s)); memset(right,0,sizeof(right)); l=0; num=0; for(i=1;i<=n;i++) { scanf("%d",&p[i]); if(i==1) right[i]=p[i]+1; else right[i]=right[i-1]+(p[i]-p[i-1])+1; } for(i=1;i<=n;i++) s[right[i]]=')'; w[right[n]]=dfs(1,right[n]); for(i=1;i<=right[n];i++) { if(w[i]) ans[num++]=w[i]; } if(num==0) printf("/n"); for(i=0;i<num-1;i++) printf("%d ",ans[i]); printf("%d/n",ans[num-1]); } return 0; }