uva 11997K Smallest Sums

原题：
You’re given k arrays, each array has k integers. There are k k ways to pick exactly one element in each array and calculate the sum of the integers. Your task is to find the k smallest sums among them.
Input
There will be several test cases. The first line of each case contains an integer k (2 ≤ k ≤ 750). Each of the following k lines contains k positive integers in each array. Each of these integers does not exceed 1,000,000. The input is terminated by end-of-file (EOF).
Output
For each test case, print the k smallest sums, in ascending order.
Sample Input
3
1 8 5
9 2 5
10 7 6
2
1 1
1 2
Sample Output
9 10 12
2 2

中文：
给你k个数组，每个数组有k个数。现在问你每个数组中取一个数加一起求和，那么最小的前k个是多少？

#include <bits/stdc++.h>using namespace std;int k;vector<int> v[751],ans;struct item{    int s,b;    item(int s,int b):s(s),b(b){}    bool operator < (const item &rhs) const    {        return s>rhs.s;    }};void ini(){    for(int i=1;i<=k;i++)        v[i].clear();}void Merge(vector<int> tmp){    priority_queue<item> pq;    for(int i=0;i<ans.size();i++)        pq.push(item(ans[i]+tmp[0],0));    for(int i=0;i<k;i++)    {        item t=pq.top();        pq.pop();        ans[i]=t.s;        int b=t.b;        if(b+1<k)            pq.push(item(t.s-tmp[b]+tmp[b+1],b+1));    }}int main(){    ios::sync_with_stdio(false);    while(cin>>k)    {        ini();        for(int i=1;i<=k;i++)        {            for(int j=1;j<=k;j++)            {                int res;                cin>>res;                v[i].push_back(res);            }            sort(v[i].begin(),v[i].end());        }        ans=v[1];        for(int i=2;i<=k;i++)            Merge(v[i]);        for(int i=0;i<ans.size();i++)            if(i!=ans.size()-1)                cout<<ans[i]<<" ";            else                cout<<ans[i]<<endl;    }    return 0;}

思路：
训练指南中的例题，想了一会没思路，直接看的答案。想法很巧妙。
假设有两个数组求前k个和最小保存在ans当中，现在有第三个数组，那么三个数组的和的最小值就是把当前保存的前k个最小值加上第三个数组当中的k个值即可。
用优先队列维护前k个最小值，然后用递推的思想每次算出加上第三个数组中第i个值的和即可。