Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12948    Accepted Submission(s): 7412 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9380    Accepted Submission(s): 5481 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1027 Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10388    Accepted Submission(s): 5978 Problem Description Now our…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9458    Accepted Submission(s): 5532 Problem Description Now our hero finds the door to the BEelzebub feng5166. He op…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4865    Accepted Submission(s): 2929 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4865    Accepted Submission(s): 2929 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…

第 m 大的 n 个数全排列 DFS可过 #include <iostream> using namespace std; int n,m; ]; bool flag; ]; void dfs(int x) { if(x>n){ m--; ; return ; } if(!flag){ ;i<=n;i++) { if(!vis[i]&& !flag) { vis[i]=; ans[x]=i; dfs(x+); vis[i]=; } } } } int main()…

这道题目最开始完全不懂,后来百度了一下,原来是字典序.而且还是组合数学里的东西.看字典序的算法看了半天才搞清楚,自己仔细想了想,确实也是那么回事儿.对于长度为n的数组a,算法如下:(1)从右向左扫描,找到满足a[i]<a[i+1]的第一个i,也就是i = max{i|a[i]<a[i+1]},同时也意味着a[i+1]~a[n]是升序:(2)从右向左扫描,找到满足a[j]>a[i]的第一个j,也就是j = max{j|a[j]>a[i]},a[j]也是满足大于a[i]的最小数:(3)…

直接选择序列的方法解本题,可是最坏时间效率是O(n*n),故此不能达到0MS. 使用删除优化,那么就能够达到0MS了. 删除优化就是当须要删除数组中的元素为第一个元素的时候,那么就直接移动数组的头指针就能够了,那么时间效率就是O(1)了,而普通的删除那么时间效率是O(n),故此大大优化了程序. 怎样直接选择第k个序列,能够參考本博客的Leetcode题解.Leetcode题有个一模一样的题目.只是没有使用删除优化. 看见本题的讨论中基本上都是使用STL解,还有沾沾自喜的家伙,只是使用STL解决本…

解题报告:1-n这n个数,有n!中不同的排列,将这n!个数列按照字典序排序,输出第m个数列. 第一次TLE了,没注意到题目上的n和m的范围,n的范围是小于1000的,然后m的范围是小于10000的,很明显,如果暴力枚举的话,时间复杂度就是O(n*m),为10^7,所以可以通过.这题其实就是一个排列生成的题,排列生成有多种方法,这题的关键是当找到答案的时候,要及时退出暴力过程,要不然肯定TLE. #include<cstdio> #include<iostream> #include…