UVa11292-Dragon of Loowater

用两个优先队列维护龙头的直径和骑士的高度，不断从队列中抛出元素，第一个满足条件的即为最小的。

#include 
#include 

using namespace std;

int main(int argc, char const *argv[]) {
    int n, m;
    while (scanf("%d%d", &n, &m) == 2 && n + m) {
        priority_queue<int, vector<int>, greater<int> > head, knight;
        for (int i = 0; i < n; i++) {
            int dia;
            scanf("%d", &dia);
            head.push(dia);
        }
        for (int i = 0; i < m; i++) {
            int hei;
            scanf("%d", &hei);
            knight.push(hei);
        }
        int sum = 0;
        bool ok = true;
        while (!head.empty()) {
            int dia = head.top();
            while (!knight.empty() && knight.top() < dia) {
                knight.pop();
            }
            if (knight.empty()) {
                ok = false;
                break;
            }
            sum += knight.top();
            knight.pop();
            head.pop();
        }
        if (ok) {
            printf("%d\n", sum);
        } else {
            puts("Loowater is doomed!");
        }
    }
    return 0;
}