poj 2442 Sequence（贪心，堆）

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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

12 31 2 32 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;}

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