POJ 2442 Sequence

Sequence Time Limit: 6000MS   Memory Limit: 65536K Total Submissions: 6120   Accepted: 1897 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

 

m个有序表的前n个最小和可由m-1个有序表的qian那个最小和与第m个有序表形成。以此类推，可先参考此题：http://www.cnblogs.com/jackge/archive/2013/04/28/3049766.html

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>

using namespace std;

const int N=2010;

int m,n,a[N],b[N],c[N],d[N];

struct node{
    int id,x;
    bool operator < (const node &a) const{
        return a.x<x;
    }
};

void solve(){
    for(int i=0;i<n;i++)
        d[i]=0;
    //priority_queue<node,vector<node>,cmp> q;
    priority_queue<node> q;
    node cur;
    for(int i=0;i<n;i++){
        cur.id=i;
        cur.x=a[i]+b[d[i]];
        q.push(cur);
    }
    int k=n,cnt=0;
    while(k--){
        cur=q.top();
        q.pop();
        c[cnt++]=cur.x;
        cur.x=a[cur.id]+b[++d[cur.id]];
        q.push(cur);
    }
    for(int i=0;i<n;i++)
        a[i]=c[i];
}

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&m,&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a,a+n);
        for(int i=1;i<m;i++){
            for(int j=0;j<n;j++)
                scanf("%d",&b[j]);
            sort(b,b+n);
            solve();
        }
        for(int i=0;i<n-1;i++)
            printf("%d ",a[i]);
        printf("%d\n",a[n-1]);
    }
    return 0;
}