UVa11997 K Smallest Sums  题目: K Smallest Sums You're given k arrays, each array has k integers. There are kk ways to pick exactly one element in each array and calculate the sum of the integers. Your task is to find the k smallest sums among them. In…

UVA - 11997 K Smallest Sums Time Limit: 1000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status] Description   Problem K K Smallest Sums You're given k arrays, each array has k integers. There are kk ways to pick…

#include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<algorithm> #include<stack> #in…

思路 经典的k路归并问题 问题先转换为2路的有序表归并 先让A[1~k]都和B[1]相加,然后加入堆中,取出堆顶(A[x]+B[y])之后,再放入A[x]+B[y+1] 代码 #include <cstdio> #include <cstring> #include <algorithm> #include <queue> using namespace std; struct QNode{ int val,s; bool operator < (co…

题面 背景 输入 输出 翻译(渣自翻) 给定K个包含K个数字的表,要求将其能产生的\( k^{k} \)个值中最小的K个输出出来 题解 k路归并问题的经典问题 可以转化为二路归并问题求解 考虑A[],B[]两个有序数组 使用堆,记录一些二元组\( (x,y) \),x表示值,y表示对应的b的下标,因为我们是把b合并到a上,所以我们能够根据记录的下标推出后面的值 然后两两合并 所以就很简单 放代码 #include <cstdio> #include <algorithm> #inc…

Problem K K Smallest Sums You're given k arrays, each array has k integers. There are kk ways to pick exactly one element in each array and calculate the sum of the integers. Your task is to find the k smallest sums among them. Input There will be se…

11997 - K Smallest Sums You’re given k arrays, each array has k integers. There are kk ways to pick exactly one element in eacharray and calculate the sum of the integers. Your task is to find the k smallest sums among them.InputThere will be several…

UVA - 11997 id=18702" target="_blank" style="color:blue; text-decoration:none">K Smallest Sums Time Limit: 1000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Submit Status You're given k arrays, each array has k…

传送门 Description Input Output Translation · 给定k个长度为k的数组,把每个数组选一个元素加起来,这样共有kk种可能的答案,求最小的k个 Sample Input Sample Output Hint k<=750 Solution 显然可以一行一行做,同时如果S+now[j]最小,需要S最小.即我们只需要记录前i行的最小的k个ans,分别与当前行的k个相加,从k2个ans中选择前k个小的记录.显然前k小的可以开一个大根堆维护.即如果size>k则pop…

uva11997:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=24&problem=3148&mosmsg=Submission+received+with+ID+13942033 题意:给你k个数组,每个数组里有k个数,然后每个数组里面取出一个数相加会得到一个数,让你求出最小的看的数. 题解:白书的分析: 分析:这题有…