(DS 《算法竞赛入门经典》)UVA 11997 K Smallest Sums

题目大意：有k个数组，每个数组选取一个数，组成k^k个数。在这k^k个数中选择最小的前k个数

解题思路：

1、如果只有k个数组，那么最后得到的最小的前k个数应该可以由前两个数组得到的最小k个数与第三个数组

按规则运算后得到。

2、如果每个数组只有3个数.那么前两个数组(a:(a0,a1,a2)    b:(b0,b1,b2，a与b数组都已经有序)运算后有的结果矩阵如下：

a0+b0,a0+b1,a0+b2

a1+b0,a1+b1,a1+b2

a2+b0,a2+b1,a2+b2

在这个矩阵中，a0+b0肯定是最小的。如果一个队列里只维持3个最小数，那么下一个产生最小数的位置是

a0+b1.分析如下：

由a1+b1>a1+b0可知，a1+b1及其后面的不可能产生下一个最小数.依照此逻辑可知下一个可能产生最小数的位置

是a0+b1

3、在以下的这种解法中，他并没有算出k^k个结果，它其实质算了n+n个可能产生最小数的结果而已

。

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

31 8 59 2 510 7 621 11 2

Output for the Sample Input

9 10 122 2

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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!

代码如下：

/* * UVA_11997.cpp * *  Created on: 2014年12月30日 *      Author: Administrator */#include <iostream>#include <cstdio>#include <queue>#include <algorithm>using namespace std;const int maxn = 758;int A[maxn][maxn];struct Item{int s,b;Item(int s,int b):s(s),b(b){}bool operator < (const Item& b) const{return s > b.s;}};void merge(int* A,int* B,int* C,int n){priority_queue<Item> q;int i;for(i = 0 ; i < n ; ++i){q.push(Item(A[i] + B[0],0));}for(i = 0 ; i < n ; ++i){Item item = q.top();q.pop();C[i] = item.s;int b = item.b;if(b+1 < n){q.push(Item(item.s - B[b] + B[b+1],b+1));}}}int main(){int n;while(scanf("%d",&n) == 1){int i;int j;for(i = 0 ; i < n ; ++i){for(j = 0 ; j < n ; ++j){scanf("%d",&A[i][j]);}sort(A[i],A[i]+n);}for(i = 1 ; i < n ; ++i){merge(A[0],A[i],A[0],n);}printf("%d",A[0][0]);for(i = 1 ; i < n ; ++i){printf(" %d",A[0][i]);}printf("\n");}return 0;}