摘要:
题意: "传送门" 题解: dp+线段树 O(nlogn)最长上升子序列 include include include include include include define ll long long define ls x'9')) ch=getchar(); if(ch==' ') o= 阅读全文
摘要:
题意: "传送门" 题解: 由于最多只有两件物品,所以转移主件的时候暴力转移附件即可 阅读全文
摘要:
题意: "传送门" 题解: dp 可以看做是从起点出发走两条不同的路径到终点,设dp[i][j][k][l],注意两个点(即前两维和后两维)要同时转移才能保证不走重复路径 include include include include include include define ll long l 阅读全文
摘要:
题意: "传送门" 题解: 区间dp 断环为链,由于是环,要枚举从哪个数开始合并。 2,3,5,10可以看做是2,3,5,10,2这段区间的合并,然后直接最简单的区间dp即可 注意:i要逆序枚举 include include define N 500 using namespace std; in 阅读全文