Ultraviolet A 11997-K smallest sums (priority queue + multi-channel merge)

UVA 11997 - K Smallest Sums

题目链接

题意：给定k个数组，每一个数组k个数字，要求每一个数字选出一个数字，构成和，这样一共同拥有kk种情况，要求输出最小的k个和

思路：事实上仅仅要能求出2组的前k个值，然后不断两两合并就能够了，由于对于每两组，最后答案肯定是拿前k小的去组合。然后问题就变成怎么求2组下的情况了，利用一个优先队列维护，和作为优先级，先把原数组都从小到大排序好了，那么先把a[i] + b[0] (i < k)入队，然后往后取k个，每次取完之后，拿最小那个在把b往后推一个，这样保证每次队列中都会有当前的最小值存在，利用优先队列取出就可以，总复杂度为O(k2log(k))

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int N = 755;int n, a[N][N];struct State {    int b, sum;    State() {}    State(int b, int sum) {this->b = b;this->sum = sum;    }    bool operator < (const State& c) const {return sum > c.sum;    }};void merge(int *a, int * b) {    priority_queue<State> Q;    for (int i = 0; i < n; i++)Q.push(State(0, a[i] + b[0]));    for (int i = 0; i < n; i++) {State now = Q.top();a[i] = now.sum;Q.pop();Q.push(State(now.b + 1, now.sum - b[now.b] + b[now.b + 1]));    }}int main() {    while (~scanf("%d", &n)) {for (int i = 0; i < n; i++) {    for (int j = 0; j < n; j++)scanf("%d", &a[i][j]);    sort(a[i], a[i] + n);}for (int i = 1; i < n; i++)    merge(a[0], a[i]);for (int i = 0; i < n - 1; i++)    printf("%d ", a[0][i]);printf("%d\n", a[0][n - 1]);    }    return 0;}

UVA 11997 - K Smallest Sums(优先队列+多路合并)