hzau 1209 Deadline（贪心）

K．Deadline There are N bugs to be repaired and some engineers whose abilities are roughly equal. And an engineer can repair a bug per day. Each bug has a deadline A[i].

Question: How many engineers can repair all bugs before those deadlines at least? 1<=n<= 1e6. 1<=a[i] <=1e9

Input Description There are multiply test cases. In each case, the first line is an integer N , indicates the number of bugs. The next line is n integers indicates the deadlines of those bugs.

Output Description There are one number indicates the answer to the question in a line for each case.

Input 4 1 2 3 4

Output 1

排序，大于N的不用考虑，否则超时

贪心

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 1e6 + 5;

int a[MAXN];
int b[MAXN];
int tot;

int main()
{
    int n;
    int i;
    int j;
    int maxB;
    int lMaxB;

    while (~scanf("%d", &n)) {
        int n2 = n;
        for (i = 0; i < n; ++i) {
            scanf("%d", &a[i]);
            if (a[i] >= n2) {
                --i;
                --n;
            }
        }
        //cout << n << endl;
        sort(a, a + n);

        tot = 0;
        b[tot++] = 1;
        for (i = 1; i < n; ++i) {
            maxB = -1;
            for (j = 0; j < tot; ++j) {
                if (b[j] < a[i] && b[j] > maxB) {
                    maxB = b[j];
                    lMaxB = j;
                    break;
                }
            }

            if (maxB == -1) {
                b[tot++] = 1;
            } else {
                ++b[lMaxB];
            }

        }

        printf("%d
", tot);
    }

    return 0;
}

但是为什么用set写超时呢，不是应该比数组遍历查找要快吗？

超时代码：

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 1e6 + 5;

int a[MAXN];

struct cmp {
    bool operator()(int a, int b)
    {
        return a > b;
    }
};
multiset<int, cmp> st;

int main()
{
    int n;
    int i;
    multiset<int, cmp>::iterator it;
    int tmp;

    while (~scanf("%d", &n)) {
        st.clear();
        int n2 = n;
        for (i = 0; i < n; ++i) {
            scanf("%d", &a[i]);
            if (a[i] >= n2) {
                --i;
                --n;
            }
        }
        //cout << n << endl;
        sort(a, a + n);

        st.insert(1);
        for (i = 1; i < n; ++i) {
            it = st.upper_bound(a[i]);
            if (it == st.end()) {
                st.insert(1);
            } else {
                tmp = *it + 1;
                st.erase(it);
                st.insert(tmp);
            }
        }

        printf("%d
", st.size());
        
    }

    return 0;
}

其实这题可以用 任务数 / 天数，向上取整得到最少需要的人数，取最大值

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 1e6 + 5;

int maxA;
int b[MAXN];//b[i]保存当天要完成的任务数
int sum[MAXN];//sum[i]代表之前一共要完成任务数

int main()
{
    int n;
    int i;
    int a;
    int ans;

    while (~scanf("%d", &n)) {

        memset(b, 0, sizeof(b));
        for (i = 0; i < n; ++i) {
            scanf("%d", &a);
            if (a > n) {
                continue;
            }
            ++b[a];
            if (a > maxA) {
                maxA = a;
            }
        }
        //cout << n << endl;

        sum[0] = 0;
        ans = 1;//最小为1个工人...//初始化为0不对。因为有可能所有日期都大于n，那么结果应该为1，而不是0
        for (i = 1; i <= maxA; ++i) {
            sum[i] = sum[i - 1] + b[i];
            ans = max(ans, (sum[i] + i - 1) / i);
        }

        printf("%d
", ans);
    }

    return 0;
}