兩個(gè)數(shù)組，從每個(gè)數(shù)組中取一個(gè)數(shù)相加，求最大的前k個(gè)和?
比如：?
數(shù)組A：1,2,3?
數(shù)組B：4,5,6?
則最大的前2個(gè)和：9,8。?
ps：結(jié)果放到數(shù)組C[k]中?

http://www.cnblogs.com/372465774y/archive/2012/07/09/2583866.html

Sequence

Time Limit:?6000MS	?	Memory Limit:?65536K
Total Submissions:?5137	?	Accepted:?1572

Description

Given m sequences, each contains n non-negative integer. Now we may select one number from each sequence to form a sequence with m integers. It's clear that we may get n ^ m this kind of sequences. Then we can calculate the sum of numbers in each sequence, and get n ^ m values. What we need is the smallest n sums. Could you help us?

Input

The first line is an integer T, which shows the number of test cases, and then T test cases follow. The first line of each case contains two integers m, n (0 < m <= 100, 0 < n <= 2000). The following m lines indicate the m sequence respectively. No integer in the sequence is greater than 10000.

Output

For each test case, print a line with the smallest n sums in increasing order, which is separated by a space.

Sample Input

1 2 3 1 2 3 2 2 3

Sample Output

3 3 4

題意：給你n*m的矩陣，然后每行取一個(gè)元素，組成一個(gè)包含n個(gè)元素的序列，一共有n^m種序列，

讓你求出序列和最小的前n個(gè)序列的序列和。

解題步驟：

1.將第一序列讀入data1向量中，并按升序排序。

2.將數(shù)據(jù)讀入data2向量中，并按升序排序。

??? 將data2[0] + data1[i] ( 0<=i<=n-1)讀入dataq向量中

??? 建堆。

??? 然后data2[1] + data1[i] (0<=i<=n-1),如果data2[1] + data1[i]比堆dataq的頂點(diǎn)大,則退出，否則刪除

??? 堆的頂點(diǎn)，插入data2[1] + data1[i]。然后是data2[2],...data2[n - 1]

3.將dataq的數(shù)據(jù)拷貝到data1中，并對data1按升序排序

4.循環(huán)2,3步，直到所有數(shù)據(jù)讀入完畢。

5.打印data1中的數(shù)據(jù)即為結(jié)果。

[cpp]?view plaincopy

#include?<cstdio>??

#include?<cstring>??

#include?<string>??

#include?<vector>??

#include?<queue>??

#include?<algorithm>??

#include?<iostream>??

using?namespace?std;??

int?main()??

{??

????int?t;??

????int?n,m;??

????int?num1[2010];??

????int?num2[2010];??

????priority_queue<int,deque<int>,less<int>?>?big;??

????scanf("%d",&t);??

????while(t--)??

????{??

????????scanf("%d%d",&m,&n);??

????????for(int?i=0;i<n;i++)??

????????scanf("%d",&num1[i]);??

????????sort(num1,num1+n);??

????????for(int?i=1;i<m;i++)??

????????{??

????????????for(int?j=0;j<n;j++)??

????????????{??

????????????????scanf("%d",&num2[j]);??

????????????????big.push(num1[0]+num2[j]);??

????????????}??

????????????sort(num2,num2+n);??

????????????for(int?k=1;k<n;k++)??

????????????for(int?l=0;l<n;l++)??

????????????{??

????????????????if(num1[k]+num2[l]>big.top())??

????????????????????break;??

????????????????????big.pop();??

????????????????????big.push(num1[k]+num2[l]);??

????????????}??

????????????for(int?k=0;k<n;k++)??

????????????{??

????????????????num1[n-k-1]=big.top();??

????????????????big.pop();??

????????????}??

????????}??

????????printf("%d",num1[0]);??

????????for(int?i=1;i<n;i++)??

????????printf("?%d",num1[i]);??

????????puts("");??

????}??

}??

?

開始忘記給他們排序、WA了一次、郁悶！

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <vector>
#include <algorithm>
#define N 2003
using namespace std;
struct node
{
??? int i;
??? int n;
};
struct cmp
{
? bool operator ()(const node&a,const node&b)
? {
????? return a.n>b.n;
? }
};
int a[N],b[N],c[N];
int m,n;
int d[N];
void del()
{
??? int i;
??? for(i=0;i<n;i++)
???????? d[i]=0;
??? node t;
??? priority_queue<node,vector<node>,cmp> Q;
??? for(i=0;i<n;i++)
??? {
??????? t.i=i;
??????? t.n=a[i]+b[d[i]];
??????? Q.push(t);
??? }
??? int te=n;i=0;
??? while(te--)
??? {
?????? t=Q.top();Q.pop();
?????? c[i++]=t.n;
?????? t.n=a[t.i]+b[++d[t.i]];
?????? Q.push(t);
??? }
??? for(i=0;i<n;i++)
???? a[i]=c[i];
}
int main()
{
??? int T;
??? int i;
??? scanf("%d",&T);
??? while(T--)
??? {
??????? scanf("%d%d",&m,&n);
??????? for(i=0;i<n;i++)
????????? scanf("%d",&a[i]);
??????? sort(a,a+n);
??????? while(--m)
??????? {
??????????? for(i=0;i<n;i++)
????????????? scanf("%d",&b[i]);
??????????? sort(b,b+n);
??????????? del();
??????? }
??????? n--;
???? for(i=0;i<n;i++)
?????? printf("%d ",a[i]);
???? printf("%d\n",a[i]);
??? }
??? return 0;
}

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