[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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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 ", L); } else printf("%d ", 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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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 ", ans); } int main() { #ifdef local freopen("data.txt", "r", stdin); // freopen("data.txt", "w", stdout); #endif op['('] = ')'; op[')'] = '('; work(); return 0; }