标签：应该 printf one amp 排序 ber inpu each ==
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
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int MAXN = 1e6 + 5; 5 6 int a[MAXN]; 7 int b[MAXN]; 8 int tot; 9 10 int main() 11 { 12 int n; 13 int i; 14 int j; 15 int maxB; 16 int lMaxB; 17 18 while (~scanf("%d", &n)) { 19 int n2 = n; 20 for (i = 0; i < n; ++i) { 21 scanf("%d", &a[i]); 22 if (a[i] >= n2) { 23 --i; 24 --n; 25 } 26 } 27 //cout << n << endl; 28 sort(a, a + n); 29 30 tot = 0; 31 b[tot++] = 1; 32 for (i = 1; i < n; ++i) { 33 maxB = -1; 34 for (j = 0; j < tot; ++j) { 35 if (b[j] < a[i] && b[j] > maxB) { 36 maxB = b[j]; 37 lMaxB = j; 38 break; 39 } 40 } 41 42 if (maxB == -1) { 43 b[tot++] = 1; 44 } else { 45 ++b[lMaxB]; 46 } 47 48 } 49 50 printf("%d\n", tot); 51 } 52 53 return 0; 54 }
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？
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 const int MAXN = 1e6 + 5; 5 6 int a[MAXN]; 7 8 struct cmp { 9 bool operator()(int a, int b) 10 { 11 return a > b; 12 } 13 }; 14 multiset<int, cmp> st; 15 16 int main() 17 { 18 int n; 19 int i; 20 multiset<int, cmp>::iterator it; 21 int tmp; 22 23 while (~scanf("%d", &n)) { 24 st.clear(); 25 int n2 = n; 26 for (i = 0; i < n; ++i) { 27 scanf("%d", &a[i]); 28 if (a[i] >= n2) { 29 --i; 30 --n; 31 } 32 } 33 //cout << n << endl; 34 sort(a, a + n); 35 36 st.insert(1); 37 for (i = 1; i < n; ++i) { 38 it = st.upper_bound(a[i]); 39 if (it == st.end()) { 40 st.insert(1); 41 } else { 42 tmp = *it + 1; 43 st.erase(it); 44 st.insert(tmp); 45 } 46 } 47 48 printf("%d\n", st.size()); 49 50 } 51 52 return 0; 53 }
标签：应该 printf one amp 排序 ber inpu each ==
原文地址：http://www.cnblogs.com/bofengyu/p/6755521.html
