网易2018校园招聘编...

[编程题]游历魔法王国

https://www.nowcoder.com/questionTerminal/f58859adc39f4edc9cd8e40ba4160339

给定一棵树，可以走L步，最多能访问多少个节点

思路：分类讨论，首先贪心走最长链，如果还有多余的步数，可以知道如果从最底端返回上面再访问节点的话，这样不是最优，应该是在最长链的某个点就走进去然后再回来，如果多了k步，那么最多可以访问k / 2个节点，因为一去一回需要2步，相当于回到这个节点

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 50 + 20;
struct Node {
    int u, v, tonext;
} e[maxn * 2];
int first[maxn], num;
void addEdge(int u, int v) {
    e[num].u = u, e[num].v = v, e[num].tonext = first[u];
    first[u] = num++;
}
int n, L;
int dp[maxn];

void dfs(int cur, int sel) {
    dp[cur] = sel;
    for (int i = first[cur]; ~i; i = e[i].tonext) {
        dfs(e[i].v, sel + 1);
    }
}

void work() {
    memset(first, -1, sizeof first);
    cin >> n >> L;
    for (int i = 0; i <= n - 2; ++i) {
        int fa;
        scanf("%d", &fa);
        addEdge(fa, i + 1);
    }
    dfs(0, 1);
    int mx = 0;
    for (int i = 0; i < n; ++i) mx = max(mx, dp[i]);
    L++;
    if (L <= mx) {
        printf("%d\n", L);
    } else printf("%d\n", min(n, mx + (L - mx) / 2));
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}

 

 

[编程题]最长公共子括号序列

https://www.nowcoder.com/questionTerminal/504ad6420b314e5bb614e1684ad46d4d

观察到答案肯定是lenstr - 1

那么暴力移除一个括号，插入到字符序列的任意位置，可以维持原长度，而且左括号 = 右括号

所以能枚举到每一种LCS = lenstr - 1的情况

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int maxn = 50 + 2;
char str[maxn], sub[maxn];
int dp[maxn][maxn];
int calc(char str[], char sub[], int lenstr) {
    memset(dp, -1, sizeof dp);
    dp[0][0] = 0;
    for (int i = 1; i <= lenstr; ++i) {
        for (int j = 1; j <= lenstr; ++j) {
            if (str[i] == sub[j]) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            }
            dp[i][j] = max(dp[i][j], dp[i - 1][j]);
            dp[i][j] = max(dp[i][j], dp[i][j - 1]);
        }
    }
    return dp[lenstr][lenstr];
}
int lenstr;
int ans;
char op[2222];
 
bool check() {
    int lef = 0;
    for (int i = 1; i <= lenstr; ++i) {
        if (sub[i] == '(') lef++;
        else lef--;
        if (lef < 0) return false;
    }
    return lef == 0;
}
 
void did(int from, int to) {
    strcpy(sub + 1, str + 1);
    if (from > to) {
        char ch = sub[from];
        for (int i = from; i >= to; --i) {
            sub[i] = sub[i - 1];
        }
        sub[to] = ch;
    } else {
        char ch = sub[from];
        for (int i = from + 1; i <= to; ++i) {
            sub[i - 1] = sub[i];
        }
        sub[to] = ch;
    }
}
map<string, bool> mp;
void work() {
    scanf("%s", str + 1);
    lenstr = strlen(str + 1);
    strcpy(sub + 1, str + 1);
    for (int i = 1; i <= lenstr; ++i) {
        for (int j = 1; j <= lenstr; ++j) {
            if (i == j) continue;
            did(i, j);
            if (check() && mp.find(sub + 1) == mp.end() && calc(str, sub, lenstr) == lenstr - 1) {
                mp[sub + 1] = true;
                ans++;
            }
        }
    }
    printf("%d\n", ans);
}
 
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    op['('] = ')';
    op[')'] = '(';
    work();
    return 0;
}