标签：区间dp
|
Language:
Brackets Sequence
Description
Let us define a regular brackets sequence in the following way:
1. Empty sequence is a regular sequence. 2. If S is a regular sequence, then (S) and [S] are both regular sequences. 3. If A and B are regular sequences, then AB is a regular sequence. For example, all of the following sequences of characters are regular brackets sequences: (), [], (()), ([]), ()[], ()[()] And all of the following character sequences are not: (, [, ), )(, ([)], ([(] Some sequence of characters ‘(‘, ‘)‘, ‘[‘, and ‘]‘ is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 ... an is called a subsequence of the string b1 b2 ... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n. Input
The input file contains at most 100 brackets (characters ‘(‘, ‘)‘, ‘[‘ and ‘]‘) that are situated on a single line without any other characters among them.
Output
Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.
Sample Input ([(] Sample Output ()[()] Source |
||||||||||
区间dp
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 305
#define INF 0x3f3f3f3f
int dp[N][N],vis[N][N];
char a[N];
inline bool judge(int i,int j)
{
if(a[i]=='('&&a[j]==')'||a[i]=='['&&a[j]==']')
return true;
return false;
}
void print(int i,int j)
{
if(i>j) return ;
if(i==j)
{
if(a[i]=='('||a[i]==')')
printf("()");
else
printf("[]");
return ;
}
if(vis[i][j]==-1)
{
printf("%c",a[i]);
print(i+1,j-1);
printf("%c",a[j]);
return ;
}
print(i,vis[i][j]);
print(vis[i][j]+1,j);
}
int main()
{
int i,j;
while(gets(a))
{
int len=strlen(a);
if(len==0)
{
printf("\n");
continue;
}
memset(dp,0,sizeof(dp));
memset(vis,-1,sizeof(vis));
for(i=0;i<len;i++)
dp[i][i]=1;
for(i=len-1;i>=0;i--)
for(j=i+1;j<len;j++)
{
dp[i][j]=dp[i+1][j]+1; //刚开始是自己匹配，即加一个
vis[i][j]=i;
if(judge(i,j)) //如果首尾匹配,这个特殊处理
{
if(dp[i][j]>dp[i+1][j-1])
{
dp[i][j]=dp[i+1][j-1];
vis[i][j]=-1;
}
}
for(int k=i+1;k<j;k++) //i~j，之间有匹配的
if(judge(i,k))
{
if(dp[i][j]>dp[i][k]+dp[k+1][j])
{
dp[i][j]=dp[i][k]+dp[k+1][j];
vis[i][j]=k;
}
}
}
print(0,len-1);
printf("\n");
}
return 0;
}
poj 1141 Brackets Sequence(区间dp)
标签：区间dp
原文地址：http://blog.csdn.net/u014737310/article/details/43906543
