codeforce 985C Liebig's Barrels(贪心+思维)
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You have m?=?n·k wooden staves. The i-th stave has length ai. You have to assemble n barrels consisting of k staves each, you can use any k staves to construct a barrel. Each stave must belong to exactly one barrel.
Let volume vj of barrel j be equal to the length of the minimal stave in it.
You want to assemble exactly n barrels with the maximal total sum of volumes. But you have to make them equal enough, so a difference between volumes of any pair of the resulting barrels must not exceed l, i.e. |vx?-?vy|?≤?l for any 1?≤?x?≤?n and 1?≤?y?≤?n.
Print maximal total sum of volumes of equal enough barrels or 0 if it‘s impossible to satisfy the condition above.
The first line contains three space-separated integers n, k and l (1?≤?n,?k?≤?105, 1?≤?n·k?≤?105, 0?≤?l?≤?109).
The second line contains m?=?n·k space-separated integers a1,?a2,?...,?am (1?≤?ai?≤?109) — lengths of staves.
Print single integer — maximal total sum of the volumes of barrels or 0 if it‘s impossible to construct exactly n barrels satisfying the condition |vx?-?vy|?≤?l for any 1?≤?x?≤?n and 1?≤?y?≤?n.
4 2 1
2 2 1 2 3 2 2 3
7
2 1 0
10 10
20
1 2 1
5 2
2
3 2 1
1 2 3 4 5 6
0
In the first example you can form the following barrels: [1,?2], [2,?2], [2,?3], [2,?3].
In the second example you can form the following barrels: [10], [10].
In the third example you can form the following barrels: [2,?5].
In the fourth example difference between volumes of barrels in any partition is at least 2 so it is impossible to make barrels equal enough.
题意:输入 n k l 你要做n个桶,每个桶需要k个木板,用木板拼好的桶相互之间体积的差距<=l,桶的体积大小就是最短的那根木板的长度大小。
第二行 共n*k个数,分别表示n*k个木板的长度。
分析:
先对边排个序
不存在的情况,就是a[n]-a[1]>l,那就是不存在,因为要是差距尽可能小,前n小的都分别作为n个桶的一块木板,那么这之中最大的差距就是a[n]-a[1],要是a[n]-a[1]都满足条件(<=l)了,那就满足条件了。
其次,要使体积和最大输出体积和,我毛想想觉得s=a[1]+……a[n],结果WA了,引起了我的深思。
因为:
eg:4 3 17
1 2 3 5 9 13 18 21 22 23 25 26
它可以这样组3组:
18 25 26
13 22 23
1 2 3
5 9 21
这样体积为1+5+13+18=37,不是简单地1 +2 +3 +5=11
所以我的思路:先要找到最大的满足条件的数,可以用二分找更快,在这组样例中,是18,它-a[1]<=l,
那么从最后开始去k-1个和18拼,s+=18,再下一个数13(25 26),再从最后找k-1个数(22 23),
再下一个数9,发现再k-1个数不够了,那就从头开始找了,(1 2 3)一组,在去(5 9 13)时,发现13
已经被取走,那就s+=5就可以了。
#include <iostream> #include<cstring> #include<string> #include<cstdio> #include<algorithm> #include<cmath> #include<deque> #include<vector> #define ll long long #define inf 0x3f3f3f3f #define mod 1000000007; using namespace std; ll a[100005]; bool cmp(ll a,ll b) { return a<b; } int main() { ll n,k,l; scanf("%I64d%I64d%I64d",&n,&k,&l); for(ll i=1;i<=n*k;i++) { scanf("%I64d",&a[i]); } sort(a+1,a+1+n*k,cmp); if(a[n]-a[1]>l) { printf("0"); } else { ll s=0; ll p=-1; for(ll i=n*k;i>=1;i--) { if(a[i]-a[1]<=l) { p=i;//找到标准数,最大的满足条件的数 break; } } s=0; int num=0;//记录从标准数向前取了多少 int j=p; for(ll i=n*k;i-(k-1)>p;i=i-(k-1))//先从后往前取 { s+=a[j--]; num++; } for(ll i=1;i<p-num+1;i=i+k)//在从前往后取 { s+=a[i]; } printf("%I64d",s); } return 0; }
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