theoinfo→Codeforces Round 1025 (Div. 2)HELP! MohammadParsaElahimanesh→Can we find each Required node in segment tree in O(1)? -739397→I’m Finally a Specialist md_faisal_ahammed→Where is div3? truth→Why les
and that was pretty much it. We were then asked to solve a problem set on Codeforces individually. I didn't understand much from the complexity proof, couldn't solve a single problem, failed to figure
CF1000E We Need More Bosses 对无向图进行缩点后求树直径即为答案。...Educational Codeforces Round 46 (Rated for Div. 2)E. We Need More Bosses 题目链接:E. We Need More Bosses 题解:tarjan有向图缩点之后求树的直径就是答案:应为在同一个强联通里的边就不是必须边,参考了这个 有个相似的题...
无向图缩点的话 在dfs的时候禁止访问父节点就好了,其他与有向图缩点一样。 然后就是求树的直径。 画图想想就理解了,从“必经之路”入手,想到环是造成非必经之路的原因。我们缩完点后得到无环连通图,即一棵树,树上每两个点之间有一条唯一的路径,路径上每条边都是必经之路。 所以只是到 tarjan缩点模板 + ...
Notably, it is the first open research to validate that reasoning capabilities of LLMs can be incentivized purely through RL, without the need for SFT. This breakthrough paves the way for future advancements in this area. We introduce our pipeline to develop DeepSeek-R1. The pipeline ...
CF1000E We Need More Bosses 题目大意: n点m边无向图,找到两个点s、t,使得s到t必须经过的边最多,求最多的必须经过边数。 思路: 题目关键在于对“必须经过的边”的理解,拿样例1来说 Copy 55512|23231/ \413---152|4 在同一个双连通分量中,任意两点都至少有两条简单路径,所以一个双连通分量内的点都...
E - We Need More Bosses CodeForces - 1000E (tarjan缩点,树的直径),"EWeNeedMoreBosses""CodeForces1000E"Yourfriendisdevelopingacomputergame.Hehasalreadydecidedhowthegameworldshouldlooklike—it
Your task is to check whether all possible strings of lengthnnnthat can be formed using the firstkkklowercase English alphabets occur as a subsequence ofsss. If the answer isNO, you also need to print a string of lengthnnnthat can be formed using the firstkkklowercase English alphabets which...
We Need More Bosses CodeForces - 1000E(缩点 建图 求桥 求直径),题意:就是求桥最多的一条路解析:先求连通分量的个数然后缩点建图求直径即可
// If ti = 1, then as a reply to the query we need to paint a blue node vi in red. // If ti = 2, then we should reply to the query by printing the shortest distance from some red node to node vi. // // It is guaranteed that the given graph is a tree and ...