Sequence 分类： 栈和队列 2015-08-05 10:10 2人阅读 评论(0) 收藏

Sequence
Time Limit: 6000MS Memory Limit: 65536K
Total Submissions: 8277 Accepted: 2708

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
题意:给你一个m*n的矩阵,从每一行中选一个元素,组成m个元素的和,问和最小的n个数;
方法:用的优先队列,优先队列中保存n个元素是前i-1个行中和最小的n个数,队头是和最大的,然后让这n个和与第i行的元素从小到大相加,如果大于队头元素,则说明队头不是最小的,则去掉队头元素,加入更小的,直到将m行都加完,队列中的n个元素就是最小的你个数.

#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-9
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define CRR fclose(stdin)
#define CWW fclose(stdout)
#define RR freopen("input.txt","r",stdin)
#pragma comment(linker, "/STACK:102400000")
#define WW freopen("output.txt","w",stdout)

const int MAX = 6000000+5;
int Arr[110][2110];
int b[2110];
priority_queue<int >Q;
int main()
{
    int T;
    int n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&m);
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
                scanf("%d",&Arr[i][j]);
            sort(Arr[i],Arr[i]+m);//进行排序
        }
        for(int i=0; i<m; i++)
        {
            Q.push(Arr[0][i]);//以第一行为基准
        }
        for(int i=1; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                b[j]=Q.top();//先用数组储存起来
                Q.pop();
            }
            for(int j=0; j<m; j++)//Arr[i]从小到大
            {
                for(int k=m-1; k>=0; k--)//b[]从小到大
                {
                    if(j==0)
                    {
                        Q.push(Arr[i][j]+b[k]);//先是Arr[i][j]与前i-1个最小的n个数相加
                    }
                    else
                    {
                        if(Arr[i][j]+b[k]<Q.top())//如果小于队头,说明队头不是n个最小的和,去掉,加入更小的
                        {
                            Q.pop();
                            Q.push(Arr[i][j]+b[k]);
                        }
                        else//跳出,因为后面的更大,就没有必要判断
                        {
                            break;

                        }
                    }
                }
            }
        }
        for(int i=m-1; i>=0; i--)
        {
            b[i]=Q.top();
            Q.pop();
        }
        for(int i=0; i<m; i++)
        {
            if(i)
                printf(" ");
            printf("%d",b[i]);
        }
        printf("\n");
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。