• UVA


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

      

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  • 原文地址:https://www.cnblogs.com/tyty-Somnuspoppy/p/7359676.html
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