HDU 4546 比赛难度
思路:由于m不是很大,如果用一个优先队列维护,如果每次能保证加入的值是最小的,那么只需要加入m次就能完成了,时间复杂度足够,那么如何保证呢,就把数列排序,维护优先队列为当前和加下一个位置和的最小值,每次一个出队,把下一个位置取于不取在入队,最后求出答案即可
代码:
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int N = 10005;
int t, n, m, a[N];
struct Node {
int sum, nextsum, nextid;
Node() {}
Node(int sum, int nextsum, int nextid) {
this->sum = sum;
this->nextsum = nextsum;
this->nextid = nextid;
}
bool operator < (const Node &c) const {
return nextsum > c.nextsum;
}
};
const int INF = 0x3f3f3f3f;
int main() {
int cas = 0;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
sort(a, a + n);
priority_queue<Node> Q;
Q.push(Node(0, a[0], 0));
int cnt = 0;
while (!Q.empty()) {
Node u = Q.top();
Q.pop();
if (u.nextid >= n) continue;
Q.push(Node(u.sum, u.sum + a[u.nextid + 1], u.nextid + 1));
Q.push(Node(u.nextsum, u.nextsum + a[u.nextid + 1], u.nextid + 1));
cnt++;
if (cnt == m) {
printf("Case #%d: %d\n", ++cas, u.nextsum);
break;
}
}
}
return 0;
}