【暑假】[实用数据结构]UVa11997 K Smallest Sums

UVa11997 K Smallest Sums

题目：

K Smallest Sums

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). The size of input file does not exceed 5MB.

Output

For each test case, print the k smallest sums, in ascending order.

Sample Input

Output for the Sample Input

9 10 12

2 2

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思路：

先思考2个有序表（sort后）AB合成的情况。组织如下

【表】1： A₁+B₁<= A₁+B₂ <= A₁+B₃ <=......

【表】2：A₂+B₁<= A₂+B₂ <= A₂+B₃ <=......

......

【表】n：A_n+B₁<= A_n+B₂ <= A_n+B₃ <=......

转化为多路归并问题且有总数限制为k，因此将n个【表】合并为一个有序【表】C且表中数据数目为k。优先队列处理：初始化优先队列为n个表的第一元素，每次在队列中取出最小元素加入C并将该最小元素在【表】中的后一个元素入队。共取k个元素。

对于k个有序表的情况：两两合并。

于是有代码如下：

 #include<cstdio>

 #include<algorithm>

 #include<queue>

 #define FOR(a,b,c) for(int a=(b);a<(c);a++)

 using namespace std;

 const int maxn =  + ;

 struct Node{

     int s,b;

     bool operator <(const Node& rhs) const{ //相反定义 <

       return s>rhs.s;

     }

 };

 void merge(int* A,int* B,int*C,int n){ //合并AB数组取其前n小的和放入C

 priority_queue<Node> Q;

   FOR(i,,n) Q.push(Node{A[i]+B[],});

   FOR(i,,n){

       Node u=Q.top();Q.pop();

       C[i]=u.s;

       int b=u.b;

       if(b+ < n) Q.push(Node{u.s-B[b]+B[b+],b+}); //加入A[a][b+1]

   }

 }

 int main(){

 int n;

 int A[maxn],B[maxn];

   while(scanf("%d",&n)==){

       FOR(j,,n) scanf("%d",&A[j]); sort(A,A+n);

       FOR(i,,n){

            FOR(j,,n) scanf("%d",&B[j]);

            sort(B,B+n);

          merge(A,B,A,n);

     }

     printf("%d",A[]);

     FOR(i,,n)

       printf(" %d",A[i]);

     printf("\n");

  }

  return ;

 }

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