快速切题 poj1068
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 19716 | Accepted: 11910 |
Description
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
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
题目大意:给你一组匹配的左右括号,定义序列P是第i个右括号前面的左括号数,定义序列W是第i个右括号到与它匹配的左括号中间的右括号(包括这个括号自身)的个数,已知P,求W
解题思路:定义ind数组分别表示第i个括号是第k个左/右括号,定义l数组表示第i个左括号前面有多少个右括号,r数组则是表示第i个右括号前面有多少个左括号,相减即可,对于括号匹配直接用的遍历,这题n如果是10e9就会有趣很多,这个时候右括号一定是匹配未被用到的最大值,也就是说其实不用遍历
应用时:5min
实际用时: 41min(读题太差)
#include<cstdio>
#include <cstring>
using namespace std;
int ind[];
bool used[];
int r[];
int l[];
int len,n,llen;
int w[];
int main(){
int t;
scanf("%d",&t);
while(t--){
memset(used,,sizeof(used));
len=,llen=;
scanf("%d",&n);
for(int i=;i<n;i++){
scanf("%d",r+i);
for(int j=;j<r[i]-(i>?r[i-]:);j++){
ind[len++]=llen;
l[llen++]=i;
}
ind[len++]=i;
}
for(int i=;i<n;i++){
int tind=r[i]+i;
used[tind]=true;
for(int j=tind-;j>=;j--){
if(!used[j]){
w[i]=i-l[ind[j]]+;
used[j]=true;
break;
}
}
}
for(int i=;i<n;i++)printf("%d%c",w[i],i==n-?'\n':' ');
}
return ;
}
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