UVA - 11997 K Smallest Sums

最新推荐文章于 2019-04-16 01:24:12 发布

acm_1361677193

最新推荐文章于 2019-04-16 01:24:12 发布

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本文链接：https://blog.youkuaiyun.com/acm_1361677193/article/details/41301725

本文详细介绍了KSmallestSums算法的实现过程及应用，包括优先队列的使用和表的合并策略，旨在帮助读者理解如何在给定数组集合中找出最小k个和。

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K Smallest Sums

Time Limit: 1000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

Problem K

K Smallest Sums

You're given k arrays, each array has k integers. There are k^k 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 several test cases. The first line of each case contains an integer k (2<=k<=750). Each of the following k lines contains k positive integers in each array. Each of these integers does not exceed 1,000,000. The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.

Output

For each test case, print the k smallest sums, in ascending order.

Sample Input

Output for the Sample Input

9 10 12
2 2

Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!

分析：

优先队列的使用。涉及到表的合并。

值得好好思考。

ac代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn=750+10;
int A[maxn][maxn];

struct Item
{
int s,b;
Item(int s,int b):s(s),b(b){}//构造函数
bool operator <(const Item &rhs) const
{
return s>rhs.s;
}
};

void merge(int *A,int *B,int *C,int n)
{
priority_queue<Item> pq;
for(int i=0;i<n;i++)
pq.push(Item(A[i]+B[0],0));
for(int i=0;i<n;i++)
{
Item item=pq.top();
pq.pop();
C[i]=item.s;
int b=item.b;
// if(b<n-1)//多余。因为总数不可能超过n，在一个表中显然也不可能超过n。
pq.push(Item(item.s-B[b]+B[b+1],b+1));
}
}

int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
scanf("%d",&A[i][j]);
sort(A[i],A[i]+n);
}
for(int i=1;i<n;i++)//从1开始合并
merge(A[0],A[i],A[0],n);
printf("%d",A[0][0]);
for(int i=1;i<n;i++)
printf(" %d",A[0][i]);
printf("\n");
}
return 0;
}