ms 两个数组，从每个数组中取一个数相加，求最大的前k个和

两个数组，从每个数组中取一个数相加，求最大的前k个和
比如：
数组A：1,2,3
数组B：4,5,6
则最大的前2个和：9,8。
ps：结果放到数组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

 

 

题意：给你n*m的矩阵，然后每行取一个元素，组成一个包含n个元素的序列，一共有n^m种序列，

让你求出序列和最小的前n个序列的序列和。

解题步骤：

1.将第一序列读入data1向量中，并按升序排序。

2.将数据读入data2向量中，并按升序排序。

将data2[0] + data1[i] ( 0<=i<=n-1)读入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的顶点大,则退出，否则删除

堆的顶点，插入data2[1] + data1[i]。然后是data2[2],...data2[n - 1]

3.将dataq的数据拷贝到data1中，并对data1按升序排序

4.循环2,3步，直到所有数据读入完毕。

5.打印data1中的数据即为结果。

#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#include<queue>
#include<algorithm>
#include<iostream>
usingnamespacestd;
intmain()
{
intt;
intn,m;
intnum1[2010];
intnum2[2010];
priority_queue<int,deque<int>,less<int>>big;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&m,&n);
for(inti=0;i<n;i++)
scanf("%d",&num1[i]);
sort(num1,num1+n);
for(inti=1;i<m;i++)
{
for(intj=0;j<n;j++)
{
scanf("%d",&num2[j]);
big.push(num1[0]+num2[j]);
}
sort(num2,num2+n);
for(intk=1;k<n;k++)
for(intl=0;l<n;l++)
{
if(num1[k]+num2[l]>big.top())
break;
big.pop();
big.push(num1[k]+num2[l]);
}
for(intk=0;k<n;k++)
{
num1[n-k-1]=big.top();
big.pop();
}
}
printf("%d",num1[0]);
for(inti=1;i<n;i++)
printf("%d",num1[i]);
puts("");
}
}

 

 

 

开始忘记给他们排序、WA了一次、郁闷！

 

#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;
}