洛谷 2921 记忆化搜索 tarjan 基环外向树

洛谷 2921 记忆化搜索 tarjan

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传送门 (https://www.luogu.org/problem/show?pid=2921)

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做这题的经历有点玄学，，起因是某个random题的同学突然发现了一个0提交0通过的题目，然后就引发了整个机房的兴趣，，然后，，就变成了16提交7通过，，

初看上去这题目就是记忆化搜索，但是环的存在使得普通的记忆化会导致漏解，继续观察发现整张图为n个点n条边，即是多个基环外向树，使用tarjan找到图中的环，显然可知，对于环上一点，能取到的最大值是环的长度，对于环外一点，能取到的最大值是它走到环的长度加上环长，之后采用记忆化搜索或dp即可得解

Warning:

1. 从样例可以显然发现，存在自环
2. 开始写tarjan时错误的理解了low数组的含义，将其与from数组混淆

#include <cstdio>
#include <cstring>
#include <algorithm>const int maxn = 100000 + 100;
int next[maxn];
int dfn[maxn], low[maxn], size[maxn], sta[maxn];
int from[maxn];
int vis[maxn];
int stackTop = 0;
int tim = 0;
int n;
int ans[maxn];void tarjan(int x) {tim++;stackTop++;sta[stackTop] = x;dfn[x] = low[x] = tim;vis[x] = 1;if (!dfn[next[x]]) {tarjan(next[x]);low[x] = std :: min(low[next[x]], low[x]);} else if (vis[next[x]]) {low[x] = std :: min(low[x], dfn[next[x]]);}if (low[x] == dfn[x]) {while (sta[stackTop] != x) {size[x]++;from[sta[stackTop]] = x;vis[sta[stackTop]] = 0;stackTop--;}vis[x] = 0;stackTop--;size[x]++;from[x] = x;}
}void dfs(int x) {if (ans[x] > 0) return;if (from[x] != x || size[x] > 1) {ans[x] = size[from[x]];return;} else if (next[x] == x) {ans[x] = 1;return;} else {dfs(next[x]);ans[x] = 1 + ans[next[x]];}
}int main () {scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d", &next[i]);}for (int i = 1; i <= n; i++) {if (!dfn[i]) tarjan(i);}//for (int i = 1; i <= n; i++) //     printf("%d\n", from[i]);for (int i = 1; i <= n; i++)if (ans[i] == 0) dfs(i);for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);return 0;
}