poj 2442 Sequence stl'练习

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

#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <stdlib.h>#include <math.h>#include <ctype.h>#include <queue>#include <map>using namespace std;int main(){    priority_queue<int>q;    int t,i,j,k;    int n,m;    int a[110][2050];    int b[2050];    scanf("%d",&t);    while(t--)    {        scanf("%d %d",&m,&n);        for(i = 0; i < m; i++)        {            for(j = 0; j < n; j++)            {                scanf("%d",&a[i][j]);            }            sort(a[i],a[i]+n);        }       for(i = 0 ; i < n ; i++)            q.push(a[0][i]);        for(i = 1; i < m; i++)        {            for(j = 0; j < n; j++)            {                b[j] = q.top();                q.pop();            }            for(j=0; j<n; j++)            {                for(k=n-1; k>=0; k--)                {                    if(j==0)                    {                        q.push(a[i][j]+b[k]);                    }                    else                    {                        int he=a[i][j]+b[k];                        int bi=q.top();                        if(bi>he)                        {                            q.pop();                            q.push(he);                        }                    }                }            }        }        for(j = 0; j < n; j++)        {            b[j] = q.top();            q.pop();        }        printf("%d",b[n-1]);        for(i = n-2; i>=0; i--)            printf(" %d",b[i]);        printf("\n");    }    return 0;}

下面是用堆来写

male_heap就是构造一棵树，使得每个父结点均大于等于其子女结点 

pop_heap不是删除某个元素而是把第一个和最后一个元素对调后[first,end-1]进行构树，最后一个不进行构树 

push_heap是对容器内的元素再进行排序形成一个新的堆

#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>#define Max 2008using namespace std;int sum[Max],data[Max],Q[Max];int main(){    int i,j,k,t,m,n,temp;    scanf("%d",&t);    while(t--)    {        memset(sum,0,sizeof(sum));        scanf("%d %d",&m,&n);        for(i=0; i<n; i++)            scanf("%d",&sum[i]);        for(i=1; i<m; i++)        {            memset(data,0,sizeof(data));            //sort(sum,sum+n);            for(j=0; j<n; j++)                scanf("%d",&data[j]);            sort(data,data+n);            for(j=0; j<n; j++)                Q[j]=sum[0]+data[j];            make_heap(Q,Q+n);            for(j=1; j<n; j++)            {                for(k=0; k<n; k++)                {                    temp=sum[j]+data[k];                    if(temp>=Q[0])                        break;                    pop_heap(Q,Q+n);                    Q[n-1]=temp;                    push_heap(Q,Q+n);                }            }            for(j=n-1; j>=0; j--)                sum[j]=Q[j];        }        sort(sum,sum+n);        for(j=0; j<n-1; j++)            printf("%d ",sum[j]);        printf("%d\n",sum[n-1]);    }    return 0;}