Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. I…

To the Max Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 46788   Accepted: 24774 Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located wi…

题目链接:http://poj.org/problem?id=1050 发现这个题没有写过题解,现在补上吧,思路挺经典的. 思路就是枚举所有的连续的连续的行,比如1 2 3 4 12 23 34 45 123 234 345...然后把这些行对应列相加缩成一行,之后就是求最大子序列和了. /* ━━━━━┒ギリギリ♂ eye! ┓┏┓┏┓┃キリキリ♂ mind! ┛┗┛┗┛┃＼○/ ┓┏┓┏┓┃ / ┛┗┛┗┛┃ノ) ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗┛┃ ┓┏┓┏┓┃ ┛┗┛┗…

题意: 求一个最大子矩阵和. 思路: 枚举行区间,然后求一个最大子序列和. 贴一发挫code- #include <iostream> #include <cstdio> #include <string.h> #include <algorithm> using namespace std; typedef __int64 LL; const int N=1e2+10; int a[N][N]; int dp[N]; int main() { int n;…

To the Max Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 39913   Accepted: 21099 Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located wi…

To the Max Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 54338   Accepted: 28752 Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located wi…

To the Max Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 49351   Accepted: 26142 Description Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located wi…

常见优化 单调队列 形式 dp[i]=min{f(k)} dp[i]=max{f(k)} 要求 f(k)是关于k的函数 k的范围和i有关 转移方法 维护一个单调递增(减)的队列,可以在两头弹出元素,一头压入元素. 队列中维护的是两个值.一个是位置,这和k的范围有关系,另外一个是f(k)的值,这个用来维护单调性,当然如果f(k)的值可以利用dp值在O(1)的时间内计算出来的话队列中可以只维护一个表示位置的变量. 枚举到一个i的时候,首先判断队首元素的位置是否已经不满足k的范围了,如果不满足就将队首…

poj1050:http://poj.org/problem?id=1050 * maximum-subarray 问题的升级版本~ 本题同样是采用DP思想来做,同时有个小技巧处理:就是把二维数组看做一维数组.怎么去看呢,我们可以吧具有同样列号的数捆绑到一起,比如 a[1][1], a[2][1], a[3][1].....我们可以吧他们都看做 'a[1]'.因为最终的解是矩阵行数n中的任意一段,比如说:第p行到第q行, (1<=p<=q<=n), 我们要得到最终解,就一定要逐一枚举p,…

最大子段和 Ο(n) 的时间求出价值最大的子段 #include<cstdio> #include<iostream> using namespace std; int n,maxn; ],ans[]; int main(){ scanf("%d",&n); ;i<=n;i++){ scanf("%d",&val[i]); ans[i]=max(ans[i-]+val[i],val[i]); maxn=max(maxn,…