Sort
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1038 Accepted Submission(s): 231
Problem Description
Recently, Bob has just learnt a naive sorting algorithm: merge sort. Now, Bob receives a task from Alice.
Alice will give Bob N sorted sequences, and the i-th sequence includes ai elements. Bob need to merge all of these sequences. He can write a program, which can merge no more than k sequences in one time. The cost of a merging operation is the sum of the length of these sequences. Unfortunately, Alice allows this program to use no more than T cost. So Bob wants to know the smallest k to make the program complete in time.
Alice will give Bob N sorted sequences, and the i-th sequence includes ai elements. Bob need to merge all of these sequences. He can write a program, which can merge no more than k sequences in one time. The cost of a merging operation is the sum of the length of these sequences. Unfortunately, Alice allows this program to use no more than T cost. So Bob wants to know the smallest k to make the program complete in time.
Input
The first line of input contains an integer t0, the number of test cases. t0 test cases follow.
For each test case, the first line consists two integers N (2≤N≤100000) and T (∑Ni=1ai<T<231).
In the next line there are N integers a1,a2,a3,...,aN(∀i,0≤ai≤1000).
For each test case, the first line consists two integers N (2≤N≤100000) and T (∑Ni=1ai<T<231).
In the next line there are N integers a1,a2,a3,...,aN(∀i,0≤ai≤1000).
Output
For each test cases, output the smallest k.
Sample Input
1
5 25
1 2 3 4 5
Sample Output
3
思路：二分+哈夫曼；
首先二分枚举k，然后关键就是check的时候，首先看一组数据：12345，当k为3的时候，按照如果我们取的话会得到21，然后再看k=4，我们的设想肯定是递减的，但是，如果我们开始就取4个的答案为25，所以这样二分就不行，但是是我们的取法不对，我们应该保证后面的大的数字 竟量只取1次，那么最后k-1个数肯定被一次加完，不会有剩余，那么N个数如果从开始能选k个，那么剩下的肯定要是k-1的倍数，那么当剩下的不是k-1的倍数，我们要在前面添加0，个数就是k-1-（N-1）%（k-1）；这样就可以保证最优，也就单调了。
1 #include<stdio.h> 2 #include<algorithm> 3 #include<string.h> 4 #include<iostream> 5 #include<queue> 6 #include<stdlib.h> 7 #include<math.h> 8 #include<set> 9 using namespace std; 10 typedef long long LL; 11 queue<int>que1; 12 queue<int>que2; 13 LL ans[100005]; 14 bool check(int mid,int N,LL T); 15 int main(void) 16 { 17 int n; 18 scanf("%d",&n); 19 while(n--) 20 { 21 int N; 22 LL T; 23 scanf("%d %lld",&N,&T); 24 for(int i = 0; i < N; i++) 25 { 26 scanf("%lld",&ans[i]); 27 } 28 sort(ans,ans+N); 29 int id = 1; 30 int l = 2; 31 int r = N; 32 while(l<=r) 33 { 34 int mid = (l+r)/2; 35 if(check(mid,N,T)) 36 { 37 r = mid-1; 38 id = mid; 39 } 40 else l = mid+1; 41 } 42 printf("%d ",id); 43 } 44 return 0; 45 } 46 bool check(int mid,int N,LL T) 47 { 48 LL ask = 0; 49 while(!que1.empty()) 50 { 51 que1.pop(); 52 } 53 while(!que2.empty()) 54 { 55 que2.pop(); 56 } 57 int x = (N-1)%(mid - 1); 58 if(x!=0) 59 { 60 x = (mid-1-x)%(mid-1); 61 while(x) 62 { 63 que1.push(0); 64 x--; 65 } 66 } 67 for(int i = 0; i < N ; i++) 68 { 69 que1.push(ans[i]); 70 ask-=ans[i]; 71 } 72 while(true) 73 { 74 LL sum = 0; 75 int c = mid; 76 while(c) 77 { 78 int flag = 0; 79 if(!que1.empty()&&!que2.empty()) 80 { 81 int t = que1.front(); 82 int tt = que2.front(); 83 if(t<=tt) 84 { 85 que1.pop(); 86 } 87 else 88 { 89 t = tt; 90 que2.pop(); 91 } 92 sum+=t; 93 c--; 94 flag = 1; 95 } 96 else if(!que1.empty()) 97 { 98 int t = que1.front(); 99 que1.pop(); 100 sum+=t; 101 c--; 102 flag = 1; 103 } 104 else if(!que2.empty()) 105 { 106 int t = que2.front(); 107 que2.pop(); 108 sum+=t; 109 c--; 110 flag = 1; 111 } 112 if(!flag) 113 break; 114 } 115 que2.push(sum); 116 ask += sum; 117 if(c>0)break; 118 } 119 if(ask > T) 120 return false; 121 else return true; 122 }