HDU-6396 - Swordsman - 优先队列

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题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=6396


题目

Lawson is a magic swordsman with k kinds of magic attributes v1,v2,v3,…,vk. Now Lawson is faced with n monsters and the i-th monster also has k kinds of defensive attributes ai,1,ai,2,ai,3,…,ai,k. If v1≥ai,1 and v2≥ai,2 and v3≥ai,3 and … and vk≥ai,k, Lawson can kill the i-th monster (each monster can be killed for at most one time) and get EXP from the battle, which means vj will increase bi,j for j=1,2,3,…,k.
Now we want to know how many monsters Lawson can kill at most and how much Lawson's magic attributes can be maximized.

输入

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line has two integers n and k (1≤n≤105,1≤k≤5).
The second line has k non-negative integers (initial magic attributes) v1,v2,v3,…,vk.
For the next n lines, the i-th line contains 2k non-negative integers ai,1,ai,2,ai,3,…,ai,k,bi,1,bi,2,bi,3,…,bi,k.
It's guaranteed that all input integers are no more than 109 and for j=1,2,3,…,k.
It is guaranteed that the sum of all n ≤5×105.
The input data is very large so fast IO (like fread) is recommended.

输出

For each test case:
The first line has one integer which means the maximum number of monsters that can be killed by Lawson.
The second line has k integers and the i-th integer means maximum of the i-th magic attibute.


题意

  有n只怪兽,每只怪兽有k个防御力和k个增幅。有一个剑客,他有k个攻击力,他只有当他的k个攻击力都严格大于某只怪兽的k个防御力时,他才能杀死这只怪兽,同时他的k个攻击力会从他杀死的这只怪兽这里获得增幅,问他最多能杀死多少只怪兽,此时他的k个攻击力分别是多少。


思路

  对于$k$种防御属性,对于每一种属性建$1$个以该属性为关键字的堆,一开始将所有$monster$放入第一个堆。然后依次检查每一个堆,对于第$i$个堆的堆顶$monster$,若其第$i$个属性满足被杀死的条件,则将其弹出并放入第$i+1$个堆,循环往复直至无法移动$monster$。每个怪兽最多进堆$k$次,出堆$k$次,时间复杂度$O(k\times n\times log_2n)$


实现

#include <bits/stdc++.h>
const int maxn = int(1e5) + 7;
struct Tmp {
    int hp, index;
    bool operator < (const Tmp &tmp) const {
        return hp > tmp.hp;
    }
} buf;
typedef std::priority_queue<Tmp> priority_queue;
struct Monster {
    int hp[7], ad[7], index;
} m[maxn];
int t, n, k, ad[7];
int main() {
    for (scanf("%d", &t); t; t--) {
        scanf("%d%d", &n, &k);
        for (int i = 1; i <= k; i++) scanf("%d", ad + i);
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= k; j++) scanf("%d", m[i].hp + j);
            for (int j = 1; j <= k; j++) scanf("%d", m[i].ad + j);
            m[i].index = i;
        }
        priority_queue que[7];
        for (int i = 1; i <= n; i++) que[1].push({m[i].hp[1], i});
        int pre = 0, ans = 0;
        while (true) {
            for (int i = 1; i < k; i++) {
                while (!que[i].empty() && (buf = que[i].top(), buf.hp <= ad[i])) {
                    int index = buf.index;
                    que[i + 1].push({m[index].hp[i + 1], index});
                    que[i].pop();
                }
            }
            while (!que[k].empty() && (buf = que[k].top(), buf.hp <= ad[k])) {
                ans++;
                int index = buf.index;
                for (int i = 1; i <= k; i++) ad[i] += m[index].ad[i];
                que[k].pop();
            }
            if (ans == pre) break;
            pre = ans;
        }
        printf("%d\n", ans);
        for (int i = 1; i <= k; i++) printf("%d%c", ad[i], i == k ? '\n' : ' ');
    }
    return 0;
}
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