UVA - 1626 Brackets sequence （区间dp）

题意：给定一个串，可能空串，或由'['，']'，'('，')'组成。问使其平衡所需添加最少的字符数，并打印平衡后的串。

分析：dp[i][j]表示区间(i,j)最少需添加的字符数。

1、递推。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char s[MAXN];
int dp[MAXN][MAXN];
int len;
bool match(char a, char b){//判断是否平衡
    return (a == '(' && b == ')') || (a == '[' && b == ']');
}
void solve(){
    for(int i = 0; i < len; ++i){
        dp[i + 1][i] = 0;//l>r,该情况不需添加字符
        dp[i][i] = 1;//只有一个字符，无论是什么，都需要一个字符将其补全
    }
    for(int i = len - 2; i >= 0; --i){
        for(int j = i + 1; j < len; ++j){
            dp[i][j] = len;//len长度的串，最多就需要len个字符使其平衡，此处初始化成个最大值即可
            if(match(s[i], s[j])){//如果当前串满足(S)或[S]，则转移到S这个情况。
                dp[i][j] = min(dp[i][j], dp[i + 1][j - 1]);
            }
            for(int k = i; k < j; ++k){//在当前区间里枚举分割线
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);
            }
        }
    }
}
void print(int i, int j){
    if(i > j) return;
    if(i == j){
        if(s[i] == '(' || s[i] == ')') printf("()");
        else printf("[]");
        return;
    }
    int ans = dp[i][j];
    if(match(s[i], s[j]) && ans == dp[i + 1][j - 1]){
        printf("%c", s[i]);
        print(i + 1, j - 1);
        printf("%c", s[j]);
        return;
    }
    for(int k = i; k < j; ++k){
        if(ans == dp[i][k] + dp[k + 1][j]){
            print(i, k);
            print(k + 1, j);
            return;
        }
    }
}
int main(){
    int T;
    scanf("%d", &T);
    getchar();
    while(T--){
        gets(s);
        gets(s);
        len = strlen(s);
        solve();
        print(0, len - 1);
        printf("\n");
        if(T) printf("\n");
    }
    return 0;
}

2、记忆化搜索，更好理解些。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char s[MAXN];
int dp[MAXN][MAXN];
int len;
bool match(char a, char b){
    return (a == '(' && b == ')') || (a == '[' && b == ']');
}
int solve(int i, int j){
    if(dp[i][j] != INT_INF) return dp[i][j];
    if(i > j) return dp[i][j] = 0;
    if(i == j) return dp[i][j] = 1;
    if(match(s[i], s[j])) dp[i][j] = min(dp[i][j], solve(i + 1, j - 1));
    for(int k = i; k < j; ++k){//枚举分割线，串AB所需添加的最少字符数可转移到串A所需添加的最少字符数+串B所需添加的最少字符数
        dp[i][j] = min(dp[i][j], solve(i, k) + solve(k + 1, j));
    }
    return dp[i][j];
}
void print(int i, int j){
    if(i > j) return;
    if(i == j){
        if(s[i] == '(' || s[i] == ')') printf("()");
        else printf("[]");
        return;
    }
    int ans = dp[i][j];
    if(match(s[i], s[j]) && ans == dp[i + 1][j - 1]){
        printf("%c", s[i]);
        print(i + 1, j - 1);
        printf("%c", s[j]);
        return;
    }
    for(int k = i; k < j; ++k){
        if(ans == dp[i][k] + dp[k + 1][j]){
            print(i, k);
            print(k + 1, j);
            return;
        }
    }
}
int main(){
    int T;
    scanf("%d", &T);
    getchar();
    while(T--){
        memset(dp, INT_INF, sizeof dp);
        gets(s);
        gets(s);
        len = strlen(s);
        solve(0, len - 1);
        print(0, len - 1);
        printf("\n");
        if(T) printf("\n");
    }
    return 0;
}