POJ 2955 （区间dp）

Brackets
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
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Status

Practice

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Appoint description:
Description
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：左右括号匹配，能匹配的算两个，然后让你找到最大的匹配数
题解：区间dp，从后往前弄

#include <cstdio>#include <cstring>#define M 110char str[M];int dp[M][M];bool cmpss(char a, char b){    if(a == '(' && b == ')')    {        return true;    }    else if(a == '[' && b == ']')    {        return true;    }    return false;}int Max(int a, int b){    return a > b ? a : b;}int main(){    int n;    while(1)    {        scanf("%s", str+1);        if(!strcmp(str+1, "end"))        {            break;        }        memset(dp, 0,sizeof(dp));        n = strlen(str+1);        for(int i=n-1; i>=1; i--)//区间dp做了几道题后发现都是从后向前比很省事         {            for(int j=i+1; j<=n; j++)            {                dp[i][j] = dp[i+1][j];//一般外围都需要赋予一个初值，才会出现对比                 if(str[i] == '(' || str[i] == '[')//先行判断一下需不需要比                 {                    for(int k=i+1; k<=j; k++)                    {                        if(cmpss(str[i], str[k]))                        {                            dp[i][j] = Max(dp[i][j], dp[i+1][k-1] + dp[k+1][j] + 2);                            //i 与 j 就是两个，再加上包括的和后面的，就是应有的，对比取最大                         }                    }                }            }        }        printf("%d\n", dp[1][n]);    }    return 0;}