原題： 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： 給你k個數(shù)組，每個數(shù)組有k個數(shù)。現(xiàn)在問你每個數(shù)組中取一個數(shù)加一起求和，那么最小的前k個是多少？
#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： 訓(xùn)練指南中的例題，想了一會沒思路，直接看的答案。想法很巧妙。 假設(shè)有兩個數(shù)組求前k個和最小保存在ans當中，現(xiàn)在有第三個數(shù)組，那么三個數(shù)組的和的最小值就是把當前保存的前k個最小值加上第三個數(shù)組當中的k個值即可。 用優(yōu)先隊列維護前k個最小值，然后用遞推的思想每次算出加上第三個數(shù)組中第i個值的和即可。
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