给定一个长度为n的区间:求m段连续子区间的和 最大值(其中m段子区间互不相交) 思路: dp[i][j]: 前j个元素i个连续区间最大值 (重要 a[j]必须在最后一个区间内) 转移方程:dp[i][j]=max (dp[i][j-1],dp[i-1][t]) + a[j]  ( dp[i-1][t] 是 max ( dp[i-1[k]  1<=k<=j-1 ) 第j个元素与第j-1个元素连在一起 --->dp[i][j-1]   第j个元素单独一个区间               --…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41033    Accepted Submission(s): 14763 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

HDU 1024 Max Sum Plus Plus (动态规划) Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given…

题意:给你n个球排成一行,初始都为黑色,现在给一些操作(L,R,color),给[L,R]区间内的求染上颜色color,'w'为白,'b'为黑.问最后最长的白色区间的起点和终点的位置. 解法:先离散化,为了防止离散后错误,不仅将L,R离散,还要加入L+1,L-1,R+1,R-1一起离散,这样就绝不会有问题了.然后建线段树,线段树维护四个值: 1.col  区间颜色  0 表示黑  1 表示白  -1表示无标记 2.maxi 区间内最大白区间的长度,由于白色用1表示,所以最大白区间的长度即为区间最…

HDU 1024 题目大意:给定m和n以及n个数,求n个数的m个连续子系列的最大值,要求子序列不想交. 解题思路:<1>动态规划,定义状态dp[i][j]表示序列前j个数的i段子序列的值,其中第i个子序列包括a[j], 则max(dp[m][k]),m<=k<=n 即为所求的结果 <2>初始状态: dp[i][0] = 0, dp[0][j] = 0; <3>状态转移: 决策:a[j]自己成为一个子段,还是接在前面一个子段的后面 方程: a[j]直接接在前面…

A - Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description: Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a bra…

Max Sum Plus Plus     HDU - 1024 Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutiv…

HDU 2829 区间DP & 前缀和优化 & 四边形不等式优化 n个节点n-1条线性边,炸掉M条边也就是分为m+1个区间 问你各个区间的总策略值最少的炸法 就题目本身而言,中规中矩的区间DP问题 d p[i][j]表示前i个节点,分为j个区间的最优策略值 cost[i][j]为从i到j节点的策略值 所以dp[i][j] = min(dp[k-1][j-1] + cost[k][i] 但是复杂度太高了 可以优化的地方有: cost数组值得求取: 考虑到cost(i,j)=ΣAxAy (i≤…

Tunnel Warfare HDU 1540 区间合并+最大最小值 题意 D x是破坏这个点,Q x是表示查询以x所在的最长的连续的点的个数,R是恢复上一次破坏的点. 题解思路 参考的大佬博客 这里巧妙使用了最大值最小值来进行区间的查找.上一行是大佬的详细题解,真的很妙啊. 代码实现 #include<cstdio> #include<cstring> #include<algorithm> #include<stack> #define ls (rt&l…

HDU 1231 题目大意以及解题思路见: HDU 1003题解,此题和HDU 1003只是记录的信息不同,处理完全相同. /* HDU 1231 最大连续子序列 --- 入门DP */ #include <cstdio> #include <cstring> ]; int main() { #ifdef _LOCAL freopen("D:\\input.txt", "r", stdin); #endif int n; int maxSum…

Max Sum Plus Plus Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description:  System Crawler  (2015-09-05) Description Now I think you have got an AC in Ignatius.L's "Max Sum&…

Different GCD Subarray Query Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 221    Accepted Submission(s): 58 Problem Description This is a simple problem. The teacher gives Bob a list of probl…

You Are the One Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3348    Accepted Submission(s): 1524 Problem Description The TV shows such as You Are the One has been very popular. In order to m…

区间交 Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 849    Accepted Submission(s): 377 Problem Description 小A有一个含有n个非负整数的数列与m个区间.每个区间可以表示为li,ri. 它想选择其中k个区间, 使得这些区间的交的那些位置所对应的数的和最大. 例如样例中,选择[2,5]…

最大连续子序列 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 42367    Accepted Submission(s): 19198 Problem Description 给定K个整数的序列{ N1, N2, ..., NK },其任意连续子序列可表示为{ Ni, Ni+1, ..., Nj },其中 1 <= i <= j…

https://www.luogu.org/problem/P2894 #include<cstdio> #include<iostream> using namespace std; struct ben { int lmax,rmax,len,sum,lz; }tr[400005]; void bt(int x,int l,int r)//建树 { tr[x].lz=0;//标记清空 tr[x].sum=tr[x].len=tr[x].lmax=tr[x].rmax=r-l+1…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29942    Accepted Submission(s): 10516 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

C - 最大连续子序列 Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1231 Appoint description: Description 给定K个整数的序列{ N1, N2, ..., NK },其任意连续子序列可表示为{ Ni, Ni+1, ..., Nj },其中 1 <= i <= j <= K.最大连续子…

最大连续子序列 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 43843    Accepted Submission(s): 20002   Problem Description 给定K个整数的序列{ N1, N2, ..., NK },其任意连续子序列可表示为{ Ni, Ni+1, ..., Nj },其中 1 <= i <=…

题意:最大和连续子序列的增强版,要求从一序列中取出若干段,这些段之间不能交叉,使得和最大并输出. 分析:用dp[i][j]表示前j个数取出i段得到的最大值,那么状态转移方程为dp[i][j]=max(dp[i][j-1]+a[j],dp[i-1][k]+a[j])  i-1<=k<=j-1 这个状态转移方程表达了两种不同的选择:第一个就是第j个连在第j-1个所在的段的后面,第二个就是第j个为新的一段的第一个数字. 由于数字的个数比较大,而题目中给定的m未知,怕超内存,所以要想办法开设一维数组来…

感觉有点奇怪的是这题明明是n^2的复杂度,n=1e6竟然能过= =.应该是数据水了. dp[i][j]表示前j个数,分成i段,且最后一段的最后一个为a[j]的答案.那么转移式是:dp[i][j] = max(dp[i][j-1], max{dp[i-1][t]}) + a[j],(i-1<=t<=j-1,j-1>=i).前者表示在第i段的最后一个加上a[j],后者表示a[j]另起一段.这个dp显然是可以滚动数组的,那么空间是可以接受的.然后后者可以使用一个pre数组来记录之前的最大值.具…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 44371    Accepted Submission(s): 16084 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

AC code: #include<stdio.h> int a[100005]; int main(void) { int n,i; int sum,maxn,tem,s,e,flag; while(scanf("%d",&n)!=EOF,n) { for(i=0; i<n; i++) { scanf("%d",&a[i]); } sum=0; maxn=-10000; tem=1; flag=0; for(i=0; i<n…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23697    Accepted Submission(s): 8094 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

Max Sum Plus Plus Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 22262    Accepted Submission(s): 7484   Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. T…

Max Sum Plus PlusTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 37418    Accepted Submission(s): 13363 Problem DescriptionNow I think you have got an AC in Ignatius.L's "Max Sum" problem.…

Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutive number sequ…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21336    Accepted Submission(s): 7130 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Max Sum Plus Plus Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1024 Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to mor…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5247 找连续数 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1207    Accepted Submission(s): 435 Problem Description 小度熊拿到了一个无序的数组,对于这个数组.小度熊想知道能否找到一…