poj_1141

括号匹配，结果为匹配后的串

一、记忆化搜索(search with finger)：

从[0, strlen(s)-1] 开始搜索，缩短区间，期间记录搜索过的值，防止重复

cut[l][r]代表区间[l, r]中需要从哪里分割，为输出

串长不超过100 复杂度可以接受

// Brackets sequence
// Search with fingers

#include <iostream>
#include <string>
#include <algorithm>
#include <cstring>
#include <cstdio>

#define MAXN 100

int n, dp[MAXN+10][MAXN+10], cut[MAXN+10][MAXN+10];
char s[111];

int SearchWithFinger(int l, int r)
{
    if(l > r)
        return dp[l][r] = 0;
    if(l == r)
        return dp[l][r] = 1;
    if(dp[l][r]!=-1)
        return dp[l][r];

    int num = 1<<30;
    if( (s[l]=='('&&s[r]==')')||(s[l]=='['&&s[r]==']') )
        num = std::min(num, SearchWithFinger(l+1, r-1));
    else if(s[l]=='('||s[l]=='[')
        num = std::min(num, SearchWithFinger(l+1, r)+1);
    else if(s[r]==')'||s[r]==']')
        num = std::min(num, SearchWithFinger(l, r-1)+1);

    for(int m = l; m < r; ++m){
        int temp = SearchWithFinger(l, m) + SearchWithFinger(m+1, r);
        if(num >= temp){
            num = temp;
            cut[l][r] = m;
        }
    }

    return dp[l][r] = num;

}

void PrintOut(int l, int r)
{
    if(l > r)
        return;
    if(l == r){
        if(s[l]=='('||s[l]==')')
            printf("()");
        if(s[l]=='['||s[l]==']')
            printf("[]");
    }
    else{
        if(cut[l][r] >= 0){
            PrintOut(l, cut[l][r]);
            PrintOut(cut[l][r]+1, r);
        }
        else{
            if(s[l]=='(' ){
                printf("(");
                PrintOut(l+1, r-1);
                printf(")");
            }
            if(s[l]=='['){
                printf("[");
                PrintOut(l+1, r-1);
                printf("]");
            }
        }
    }

}

int main(int argc, char const *argv[])
{
    // freopen("in", "r", stdin);
    // freopen("out", "w", stdout);
    scanf("%d", &n);
    getchar();
    while(n--){

        getchar();
        gets(s);
    

        memset(dp, -1, sizeof(dp));
        memset(cut, -1, sizeof(cut));

        SearchWithFinger(0, strlen(s)-1);

        PrintOut(0, strlen(s)-1);

        printf("
");
        if(n)
            printf("
");
    }
    return 0;
}

二、区间DP

dp[l][r]代表从区间[l, r]的串需要添加至少多少个括号才能匹配

枚举区间长度len ： 1 to strlen(s)-1

dp[i][i] = 0 任何一个单独的括号， 都需要添加至少一个括号才能匹配

cut[l][r]意义同上

// Brackets sequence
// dynamic programming

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <iostream>

#define MAXN 100

int n, dp[MAXN+5][MAXN+5], cut[MAXN+5][MAXN+5];
char s[MAXN+10];

void PrintOut(int l, int r)
{
    if(l > r)
        return ;
    if(l == r){
        if(s[l]=='('||s[l]==')')
            printf("()");
        if(s[l]=='['||s[l]==']')
            printf("[]");
    }
    else{
        if(cut[l][r] >= 0){
            PrintOut(l, cut[l][r]);
            PrintOut(cut[l][r]+1, r);
        }
        else{
            if(s[l]=='(' ){
                printf("(");
                PrintOut(l+1, r-1);
                printf(")");
            }
            if(s[l]=='['){
                printf("[");
                PrintOut(l+1, r-1);
                printf("]");
            }
        }
    }

}

int main(int argc, char const *argv[])
{
    // freopen("in", "r", stdin);

    while(gets(s)){

        memset(dp, 0, sizeof(dp));
        memset(cut, -1, sizeof(cut));

        for(int i = 0; i < strlen(s); ++i)
            dp[i][i] = 1;

        for(int len = 1; len < strlen(s); ++len){
            for(int l = 0; l+len < strlen(s); ++l){
                int r = l+len;
                dp[l][r] = 1<<30;
                if( (s[l]=='('&&s[r]==')')||(s[l]=='['&&s[r]==']') )
                    dp[l][r] = std::min(dp[l][r], dp[l+1][r-1]);
                else if(s[l]=='('||s[l]=='[')
                    dp[l][r] = std::min(dp[l][r], dp[l+1][r]+1);
                else if(s[r]==')'||s[r]==']')
                    dp[l][r] = std::min(dp[l][r], dp[l][r-1]+1);
                for(int m = l; m < r; ++m){
                    if(dp[l][r] >= dp[l][m] + dp[m+1][r]){
                        dp[l][r] = dp[l][m] + dp[m+1][r];
                        cut[l][r] = m;
                    }
                }
            }
        }
        // printf("%d
", dp[0][strlen(s)-1]);
        PrintOut(0, strlen(s)-1);
        printf("
");
    }
    return 0;
}