poj 2442 Sequence（贪心，堆）

Sequence Time Limit: 6000MS Memory Limit: 65536KTotal Submissions: 8976 Accepted: 2996

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

题目大意; 给你m行，每行有n个数。 分别从每一行取一位数，然后加和。这样的数一定会构成m^n个。输出最小的n个即可。

动态维护前n小

用堆或优先队列都可以

heap实现：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> using namespace std; const int maxn = 2005; int a[maxn], b[maxn], sum[maxn]; int main(void) { int t, m, n; cin >> t; while(t--) { scanf("%d%d", &m, &n); for(int i = 0; i < n; i++) scanf("%d", &a[i]); for(int i = 1; i < m; i++) { sort(a, a+n); for(int j = 0; j < n; j++) scanf("%d", &b[j]); sort(b, b+n); for(int j = 0; j < n; j++) sum[j] = a[j]+b[0]; make_heap(sum, sum+n); for(int j = 1; j < n; j++) for(int k = 0; k < n; k++) { int temp = b[j]+a[k]; if(temp >= sum[0]) break; pop_heap(sum, sum+n); sum[n-1] = temp; push_heap(sum, sum+n); } for(int j = 0; j < n; j++) a[j] = sum[n-j-1]; } sort(a, a+n); for(int i = 0; i < n; i++) { if(i) printf(" "); printf("%d", a[i]); } printf("\n"); } return 0; } priority_queue实现：

#include<iostream> #include<cstdio> #include<queue> #include<algorithm> using namespace std; const int maxn = 2005; int a[maxn], b[maxn]; int main(void) { int t, m, n; cin >> t; while(t--) { priority_queue<int, vector<int>, less<int> > pq; while(!pq.empty()) pq.pop(); scanf("%d%d", &m, &n); for(int i = 0; i < n; i++) scanf("%d", &a[i]); for(int i = 1; i < m; i++) { sort(a, a+n); for(int j = 0; j < n; j++) scanf("%d", &b[j]); sort(b, b+n); for(int j = 0; j < n; j++) for(int k = 0; k < n; k++) { if(pq.size() == n && a[j]+b[k] >= pq.top()) break; else if(pq.size() < n) pq.push(a[j]+b[k]); else { pq.pop(); pq.push(a[j]+b[k]); } } int k = 0; while(!pq.empty()) { a[k++] = pq.top(); pq.pop(); } } sort(a, a+n); for(int i = 0; i < n; i++) { if(i) printf(" "); printf("%d", a[i]); } printf("\n"); } return 0; }