第一眼能看出来是个dp O($n^3$) 暴力应该很好想 dp[i][j] = $\sum_{k=1}^i [a[k] < a[i]] *dp[k][j-1]$ 发现dp[i][j] 为前面小于它的数长度为j-1的总和, 用树状数组前缀和优化一下搞成$O(n^2log_n)$ 先离散化, dp时先枚举位置,再枚举上升序列的长度 树状数组要开二维哦, 打代码的时候发现dp数组可以用树状数组直接代替(小小的空间优化) 代码: #include<iostream> #include<cs…

The Battle of Chibi Time Limit: 6000/4000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 591    Accepted Submission(s): 192 Problem Description Cao Cao made up a big army and was going to invade the whole South Chin…

C - The Battle of Chibi Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 无 Description Cao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is to have a spy in Cao…

The Battle of Chibi Time Limit: 6000/4000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2899    Accepted Submission(s): 1043 Problem Description Cao Cao made up a big army and was going to invade the whole South Ch…

The Battle of Chibi Cao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is to have a spy in Cao Cao's army. But all generals and soldiers of Cao Cao were loya…

The Battle of Chibi Time Limit: 6000/4000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others) Cao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is…

C - The Battle of Chibi Description Cao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is to have a spy in Cao Cao's army. But all generals and soldiers of C…

The Battle of Chibi Cao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is to have a spy in Cao Cao's army. But all generals and soldiers of Cao Cao were loya…

The Battle of Chibi 给出一段长度为n的序列\(\{a_i\}\),求其中长度为m的严格上升子序列个数\(mod\ 10^9+7\),\(n\leq 10^3\). 解 不难想到设\(f[i][j]\)表示以第i个位置结尾,长度为j的LSIS,因此我们有 \[f[i][j]=\sum_{k=1,a_i>a_k}^{i-1}f[k][j-1]\] 边界:\(f[i][1]=1,i=1,2,...,n\),其余为0 答案:\(\sum_{i=1}^nf[i][m]\) 注意到这是\…

题意:给你一个n个数的序列,要求从中找出含m个数的严格递增子序列,求能找出多少种不同的方案 dp[i][j]表示以第i个数结尾,形成的严格递增子序列长度为j的方案数 那么最终的答案应该就是sigma(dp[i][m]); 显然有:dp[i][j] = sigma(dp[k][j - 1]); 其中 1 <= k < i 且 a[k] < a[i]; 题目要求严格递增,这个限制怎么解决? hdu4719这道题同样是这样处理,即我们只需要从小到大dp就行了. 但是复杂度是O(n^3)的,显然…

题目链接:hdu 5542 首届CCPC的C题,比赛时一起搞了好久,最后是队友A出的,当时有试过用树状数组来优化 dp,然后今天下午也用树状数组搞了一下午,结果还是踩了和当时一样的坑:我总是把用来记录状态的 dp 数组和树状数组里的内置数组混在一起使用了,而且两重循环的顺序也反了,以至于两组数据 3 2               3 2 1 2 3     和     3 2 1 程序跑都得出了相同的结果,无语...之前做 dp 和线段树结合的题时也是傻傻地分不清,说到底就是 dp 的功力很不…

题目传送门 题意:问n长度的序列,找出长度m的上升子序列的方案数. 分析:这个问题就是问:dp[i][j] = sum (dp[i-1][k]) (1 <= k <= n, a[k] < a[j]),当前长度i一层,最后下标j一层,之前的最后下标k一层,这样是n^3的复杂度.然后树状数组可以优化j,k的复杂度,就是j扫描一遍的同时,将a[j]的信息更新到树上,那么扫描就可以用 logn时间统计出k的信息,在这之前先对a[]离散化. /***************************…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5542 题意:n个数中找m个数,使得从左到右读是上升的子序列.问一共有多少种. dp(i,j)表示取到第i个位置,长为j并且最后一个数为a(i)的方案总数. 更新就比较容易了,dp(i,j)=∑(k=1->j-1)dp(i-1,k),初始化dp(i,1)=1. #include <bits/stdc++.h> using namespace std; #define lowbit(x) x &…

题目链接:http://acm.uestc.edu.cn/#/problem/show/1217 给你一个长为n的数组,问你有多少个长度严格为m的上升子序列. dp[i][j]表示以a[i]结尾长为j的上升子序列个数.常规是三个for. 这里用树状数组优化一下,类似前缀和的处理,两个for就好了. //#pragma comment(linker, "/STACK:102400000, 102400000") #include <algorithm> #include &l…

赛后当天学长就说了树状数组,结果在一个星期后赖床时才有了一点点思路…… 因为无法提交,不确定是否正确..嗯..有错希望指出,谢谢... 嗯..已经A了..提交地址http://acm.uestc.edu.cn/#/problem/show/1217 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; /** 题意:…

题意:给定一个序列,找出长度为m的严格递增序列的个数. 思路:用dp[i][j]表示长度为i的序列以下标j结尾的总个数.三层for循环肯定超时,首先离散化,离散化之后就可以用树状数组来优化,快速查找下边在j之前,值比ary[j]小且长度为i-1 的个数 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ; ; i…

dp+树状数组优化. dp[i][j]表示以a[i]结尾,最长上升序列长度为j的方案数.dp[i][j]=sum{dp[k][j-1]} 其中k<i&&a[k]<a[i]. 离散化后,可以用1000个树状数组维护. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath>…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5542 题目大意:在n个数中找长度为m的单调上升子序列有多少种方案 题目思路:DP,离散化,树状数组优化, dp[i][j]代表大小为i的数 长度为j时的方案,状态转移方程dp[i][j]=simiga(dp[1..i-1][j-1]) 如果直接求和的话,复杂度是n^3不行 用树状数组优化求和 复杂度n^2logn n<=1000,a[i]<=1e9,所以离散化搞一下就行 //更新一下 这题在HDU…

题意 给出长度为n的序列,问这个序列中有多少个长度为m的单调递增子序列. \(1\le M\le N\le 1000\) 分析 用F[i,j]表示前j个数构成以Aj为结尾的数列中,长度为i的严格递增子序列有多少个 \[ F[i,j]=\sum_{k<j\wedge A_k<A_j}F[i-1,k] \] 复杂度\(O(mn^2)\). 观察到决策集合的变化一是只增不减,二是有一个前缀范围,用树状数组维护转移即可.时间复杂度\(O(mn\log n)\) 代码 #include<iostr…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5542 Problem DescriptionCao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is to have a spy in Cao Cao's army.…

求给定序列中长度为M的上升子序列个数.$N,M<=1000$. 很容易想到方法.$f[i,j]$表示以第$i$个数结尾,长度为$j$的满足要求子序列个数.于是转移也就写出来了$f[i][j]+=f[k][j-1]$   $(k<i且A_k<A_i)$.边界$f[0][0]=1$. 然后这是$O(N^2 M)$的.考虑优化.每次由于只从$j-1$也就是上一层状态中选在自己序号之前且比自己小的数来转移,用时间保证第一个要求,第二个要求,将每个数离散化一下,用$m$层的BIT维护以离散化值为下…

题目链接:http://acm.uestc.edu.cn/#/problem/show/1217 题目大意就是求一个序列里面长度为m的递增子序列的个数. 首先可以列出一个递推式p(len, i) = sum(p(len-1, j)) (a[j] < a[i]) p(len, i)表示以第i个结尾的长度为len的子序列的个数. 但是如果按照递增子序列的思想,然后直接弄的话,复杂度是n*m*n的. 如果需要优化的话,可以优化的地方就是那个求sum的过程. 把p数组映射到树状数组,那么求和的过程就能在…

http://acm.hdu.edu.cn/showproblem.php?pid=5542 [题意] 给定长为n的序列,问有多少个长为m的严格上升子序列? [思路] dp[i][j]表示以a[i]结尾的长度为j的严格上升子序列有多少个 dp[i][j]=sum{dp[k][j-1]},k小于i且a[k]<a[i] 区间dp,复杂度为O(n^3) 会超时,第三层循环求和用树状数组加速 时间复杂度为O(n^2logn) 离散化的时候排序后不去重(否则WA) [Accepted] #include<…

acm.hdu.edu.cn/showproblem.php?pid=5542 [Accepted] #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<cmath> using namespace std; typedef long long ll; ; ; char str[maxn]…

[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5542 [算法] 树状数组优化DP [代码] #include<bits/stdc++.h> using namespace std; #define MAXN 1010 ; int i,j,T,TC,n,m,len,ans; int a[MAXN],tmp[MAXN],rk[MAXN]; int f[MAXN][MAXN]; class BinaryIndexedTree { private…

$AcWing$ $Description$ $Sol$ 首先显然是是以严格递增子序列的长度为阶段,由于要单调递增,所以还要记录最后一位的数值 $F[i][j]$表示前$i$个数中以$A_i$结尾的长度为j单调递增序列有多少个 $F[i][j]=\sum_{k<i且A_k<A_i}^{ }F[k][j-1]$ 注意到,如果没有$A_k<A_i$这个条件我们就可以直接维护前缀和了 有$A_k<A_i$这个条件,可以考虑维护$A_i$为下标,$F[i][j-1]$为值的数组的前缀和 $…

Cao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is to have a spy in Cao Cao's army. But all generals and soldiers of Cao Cao were loyal, it's impossible t…

题意: 给你一个长度为n的数组,你需要从中找一个长度为m的严格上升子序列 问你最多能找到多少个 题解: 我们先对原序列从小到大排序,排序之后的序列就是一个上升序列 这里如果两个数相等的话,那么因为题目要我们求严格上升子序列,所以我们让这个数在数组中原来位置靠后的排序之后让它靠前(靠前也就是下标小)   我们dp方程:dp[i][j]表示截至到第i(这个i是按照没排序之前的下标)个元素,上升子序列长度为j的子序列能找到dp[i][j]个 dp转移方程:dp[i][j]=dp[1--i-1][j-1…

题目: Cao Cao made up a big army and was going to invade the whole South China. Yu Zhou was worried about it. He thought the only way to beat Cao Cao is to have a spy in Cao Cao's army. But all generals and soldiers of Cao Cao were loyal, it's impossib…

当时比赛时超时了,那时没学过树状数组,也不知道啥叫离散化(貌似好像现在也不懂).百度百科--离散化,把无限空间中无限的个体映射到有限的空间中去,以此提高算法的时空效率. 这道题是dp题,离散化和树状数组用来优化,状态转移方程:dp[i][j]=sum(dp[i-1][k])----k需要满足a[j]>a[k]&&k<j; i表示所要选的个数,j表示以第a[j]个数结尾所有的符合要求的递增串的个数,最后答案就是sum(dp[n][j])--1<=j<=p;n 为要选的…