【POJ - 2955 Brackets】 区间DP
2017-07-26 11:50
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L - Brackets
We give the following inductive definition of a “regular brackets” sequence:
the empty sequence is a regular brackets sequence,
if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
if a and b are regular brackets sequences, then ab is a regular brackets sequence.
no other sequence is a regular brackets sequence
For instance, all of the following character sequences are regular brackets sequences:
while the following character sequences are not:
Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such
that for indices i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim is
a regular brackets sequence.
Given the initial sequence
Input
The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters
The end-of-file is marked by a line containing the word “end” and should not be processed.
Output
For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.
Sample Input
Sample Output
题意:给出一个由'[' , ']' , '(' , ')' 构成的字符串,统计括号匹配的个数。
分析:一道基础的区间dp,先将dp数组初始化为0。然后遍历起点i,终点j,中间点k,要注意处理边界s[i]和s[j]是否匹配。
状态转移方程:dp[i][j] = max(dp[i][j], dp[i][k] + dp[k+1][j]);
dp[i][j] = max(dp[i][j], dp[i+1][j-1] + 2);
代码如下:
#include <set>
#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define mod 835672545
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int MX = 105;
char s[MX];
int dp[MX][MX];
int main(){
while(~scanf("%s", s)){
if(s[0] == 'e') return 0;
memset(dp, 0, sizeof(dp));
int len = strlen(s);
for(int i = len-2; i >= 0; i--){
for(int j = i+1; j <= len-1; j++){
for(int k = i; k < j; k++){
dp[i][j] = max(dp[i][j], dp[i][k] + dp[k+1][j]);
if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')){
dp[i][j] = max(dp[i][j], dp[i+1][j-1] + 2);
}
}
}
}
printf("%d\n", dp[0][len-1]);
}
return 0;
}
We give the following inductive definition of a “regular brackets” sequence:
the empty sequence is a regular brackets sequence,
if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
if a and b are regular brackets sequences, then ab is a regular brackets sequence.
no other sequence is a regular brackets sequence
For instance, all of the following character sequences are regular brackets sequences:
(), [], (()), ()[], ()[()]
while the following character sequences are not:
(, ], )(, ([)], ([(]
Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such
that for indices i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim is
a regular brackets sequence.
Given the initial sequence
([([]])], the longest regular brackets subsequence is
[([])].
Input
The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters
(,
),
[, and
]; each input test will have length between 1 and 100, inclusive.
The end-of-file is marked by a line containing the word “end” and should not be processed.
Output
For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.
Sample Input
((())) ()()() ([]]) )[)( ([][][) end
Sample Output
6 6 4 0 6
题意:给出一个由'[' , ']' , '(' , ')' 构成的字符串,统计括号匹配的个数。
分析:一道基础的区间dp,先将dp数组初始化为0。然后遍历起点i,终点j,中间点k,要注意处理边界s[i]和s[j]是否匹配。
状态转移方程:dp[i][j] = max(dp[i][j], dp[i][k] + dp[k+1][j]);
dp[i][j] = max(dp[i][j], dp[i+1][j-1] + 2);
代码如下:
#include <set>
#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <cctype>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define mod 835672545
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int MX = 105;
char s[MX];
int dp[MX][MX];
int main(){
while(~scanf("%s", s)){
if(s[0] == 'e') return 0;
memset(dp, 0, sizeof(dp));
int len = strlen(s);
for(int i = len-2; i >= 0; i--){
for(int j = i+1; j <= len-1; j++){
for(int k = i; k < j; k++){
dp[i][j] = max(dp[i][j], dp[i][k] + dp[k+1][j]);
if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')){
dp[i][j] = max(dp[i][j], dp[i+1][j-1] + 2);
}
}
}
}
printf("%d\n", dp[0][len-1]);
}
return 0;
}
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