思路：树上倍增f[i][j]=f[f[i][j-1]][j-1]还有LCA的赶脚 #include <iostream> #include <algorithm> #include <cstdio> #include <set> #include <sstream> #include<string> #include<queue> #include<stack> #include<map> #include<cmath> #include<cctype> #inc...

思路：单调栈 #include <iostream> #include <algorithm> #include <cmath> #include <ctype.h> #include <cstring> #include <cstdio> #include <set> #include <sstream> #include<string> #include<queue> #include<stack> #include<map> #include...

题意：选2个不连续长度为K的区间和的最大值思路：用sum[i]表示1到i区间中长度为K的最大值，然后从2k开始寻找2个区间长度为K和的最大值 #include <iostream> #include <algorithm> #include <cmath> #include <ctype.h> #include <cstring> #include <cstdio> #include <set> #include <sstream> #include<string> #include&l...

题意：将两个字符串合并2个字符串的字符顺序不变求最大回文串长度。思路：DP【l1】【r1】【l2】【r2】表示，字符串a【l1】【r1】和b【l2】【r2】能否组成回文串。考虑四种状态，若a[l1]==a[r1]则 dp[l1][r1][l2][r2]|=dp[l1+1][r1-1][l2][r2]若a[l1]==b[r2]则 dp[l1][r1][l2][r2]|=dp[l1+1][r1][l2][r2-1]若a[r1]==b[l2]则 dp[l1][r1][l2][r2]|=dp[l1][r1-1][l2+1][r2]若b[l2]==b[r2]则 dp[l1][r1][l2][r2]|=d...

题意：给你N个士兵。每个人的武力值为v,可组队人数为S，求怎么组队武力值最大。思路：从组队人数N-1开始，每次吧组队人数>=N的人的武力值加入，并用小顶堆维护人数的同时保证武力值最大。 #include <iostream> #include <algorithm> #include <cmath> #include <ctype.h> #include <cstring> #include <cstdio> #include <set> #include <sstream> #include...