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

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)