poj 2442 Sequence

一 原题

Sequence

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 10203		Accepted: 3352

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

Source

POJ Monthly,Guang Lin

二 分析

给定一个m行n列的整数矩阵，要求从每一行中取一个数，将得到的m个数求和，这样共有n^m种选法。输出这n^m个和中最小的n个。

维护一个ans数组，表示前k行每行取一个数能得到的最小的n个和。ans有序，利用当前ans和第(k+1)行的数num更新ans数组，更新方法就是维护一个元素个数为n的最大堆。

三 代码

Memory: 296K  Time: 1860MS  Language: C++  Result:  Accepted

/*
AUTHOR: maxkibble
LANG: C++
PROB: poj 2442 Sequence
*/

#include<iostream>
#include<queue>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxm = 105;
const int maxn = 2005;

int n, m, num[maxn], ans[maxn];
priority_queue<int> q;

void solve() {
	cin >> m >> n;
	for(int i = 0; i < n; i++) {
		cin >> num[i];
	}
	sort(num, num + n);
	memcpy(ans, num, sizeof(num));
	for(int i = 1; i < m; i++) {
		for(int i = 0; i < n; i++) {
			cin >> num[i];
		}
		sort(num, num + n);
		while(!q.empty()) q.pop();
		for(int i = 0; i < n; i++) {
			q.push(ans[0] + num[i]);
		}
		for(int i = 1; i < n; i++) {
			for(int j = 0; j < n; j++) {
				int val = ans[i] + num[j];
				if(val > q.top())
					break;
				if(val < q.top()) {
					q.pop();
					q.push(val);
				}
			}
		}
		for(int i = n - 1; i >= 0; i--) {
			ans[i] = q.top();
			q.pop();
		}
	}
	cout << ans[0];
	for(int i = 1; i < n; i++)
		cout << " " << ans[i];
	cout << endl;
}

int main() {
	ios::sync_with_stdio(false);
	int kase;
	cin >> kase;
	while(kase--) {
		solve();
	}
	return 0;
}