UV-11997-K Smallest Sums

Enter a number k (2 <= k <= 750), and then enter a matrix of k * k. The element is a positive integer not greater than 1,000,000, each row of k rows is summed up by a number, and the first k and the smallest sum are obtained. --> If the sum of k numbers is the smallest, then the two numbers in any two rows are also the smallest. Otherwise, we can find a smaller number than this value, you can calculate the k smallest sum in the two rows first, and then Merge multiple rows. [Cpp] # include <iostream> # include <algorithm> # include <queue> using namespace std; const int maxn = 750 + 10; // each row may have A maximum of 750 struct items // define the node type {int s; // s = A [I] + B [j] int B; // B = j is the subscript Item (int ss, int bb) of B [j]: s (ss), B (bb) {} bool operator <(const Item & e) const // overload operator, so that the elements in the priority queue are ranked from small to large by the value of s {return s> e. s ;}}; void merge (int * A, int * B, int k) // merge table A and table B, that is, the minimum k and {int I; priority_queue <Item> pq; for (I = 0; I <k; I ++) When k = 2 are obtained) // The first column of elements is entered into the column pq. push (Item (A [I] + B [0], 0); for (I = 0; I <k; I ++) // find the minimum k and {Item item = pq. top (); // retrieve the smallest pq in the queue. pop (); A [I] = item. s; int B = item. b; if (B + 1 <k) // if it is not the last one, add it to the column "1" to remove pq. push (Item (item. s-B [B] + B [B + 1], B + 1) ;}} int a [maxn] [maxn]; // input k * k array int main () {int k, I, j; while (cin> k) {for (I = 0; I <k; I ++) {for (j = 0; j <k; j ++) cin> a [I] [j]; sort (a [I], a [I] + k); // sorting order} for (I = 1; I <k; I ++) merge (a [0], a [I], k); // merge for (I = 0; I <K-1; I ++) // output result cout <a [0] [I] <""; cout <a [0] [k-1] <endl;} return 0 ;}