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;
我一开始的做法是用数组进行模拟这个堆的过程,但是由于每次排序函数的时间都很多,所以超时了啊、
后来看了一下题解用优先队列做的,这样判断一下如果大于队首元素就不如队列,可以省去很多的时间、、
Sequence
Time Limit: 6000MS | Memory Limit: 65536K | |
Total Submissions: 6246 | Accepted: 1945 |
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?
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.
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
#include <stdio.h> #include <string.h> #include <queue> #include <iostream> #include <stdlib.h> using namespace std; int a[40005]; int main() { int t, n, m, i, j, flat, sum; scanf("%d",&t); while(t--) { int x; scanf("%d %d",&n, &m); priority_queue<int , vector<int> , greater<int> >p; priority_queue<int , vector<int> , less<int> >q; for(i = 0; i < m; i++) { scanf("%d",&x); p.push(x); } for(i = 1; i < n; i++) { for(j = 0; j < m; j++) scanf("%d",&a[j]); while(!p.empty()) { sum = p.top(); p.pop(); for(j = 0; j < m; j++) if(q.size() < m) q.push(sum+a[j]); else if(q.size() == m && q.top() > sum+a[j]) { q.pop(); q.push(sum+a[j]); } } while(!q.empty()) { p.push(q.top()); q.pop(); } } flat = 0; for(i = 0; i < m; i++) { if(flat) printf(" "); printf("%d",p.top()); p.pop(); flat = 1; } printf("\n"); } return 0; }