题目：Problem J. Terminal
Input file: standard input
Output file: standard input
Time limit: 2 seconds
Memory limit: 256 mebibytes
N programmers from M teams are waiting at the terminal of airport. There are two shuttles at the exit
of terminal, each shuttle can carry no more than K passengers.
Now employees of the airport service need to choose one of the shuttles for each programmer. Note that:
? programmers already formed a queue before assignment to the shuttles;
? each second next programmer in the queue goes to the shuttle he or she is assigned for;
? when all programmers, assigned to the shuttle, are in, shuttle immediately closes door and leaves
the terminal;
? no two programmers from the same team may be assigned to the different shuttles;
? each programmer must be assigned to one of shuttles.
Check if its possible to find such as assignment; if the answer is positive, find minimum sum of waiting
times for each programmer. Waiting time for a person is defined as time when shuttle with this person
left the terminal; it takes one second to programmer to leave the queue and enter the assigned shuttle.
At moment 0 the first programmer begins moving to his shuttle.
Input
First line of the input contains three positive integers N, M and K (M ≤ 2000, M ≤ N ≤ 105, K ≤ 105).
Second line contains description of the queue — N space-separated integers Ai — ids of team of each
programmer in order they are placed in the queue (1 ≤ Ai ≤ M).
Output
If it is impossible to assign programmers to she shuttles following the rules above, print -1. Otherwise
print one integer — minimum sum of waiting times for all programmers.
Examples
| standard input | standard input |
| 7 3 5 2 2 1 1 1 3 1 |
39 |
| 12 3 9 1 1 1 2 3 2 2 2 2 2 2 2 |
116 |
| 2 1 2 1 1 |
4 |
思路：
如果存在可行解，那么最后一个人一定会上车，不如直接选定上第二辆车，所以第二辆车是第ｎ秒开的。
然后枚举上第一辆车的最后一个队的最后一个人是什么时候上车的。
怎么判断可行呢？０１背包即可。
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 #define MP make_pair
6 #define PB push_back
7 typedef long long LL;
8 typedef pair<int,int> PII;
9 const double eps=1e-8;
10 const double pi=acos(-1.0);
11 const int K=1e6+7;
12 const int mod=1e9+7;
13
14 int n,m,k;
15 LL ls[3000],rd[3000],cnt[3000];
16 LL ans=2e18;
17 bool dp[2][101005];
18 bool cmp(int a,int b)
19 {
20 return ls[a]<ls[b];
21 }
22 int main(void)
23 {
24 //freopen("in.txt","r",stdin);
25 scanf("%d%d%d",&n,&m,&k);
26 for(int i=1;i<=m;i++) rd[i]=i;
27 for(int i=1,x;i<=n;i++)
28 scanf("%d",&x),ls[x]=i,cnt[x]++;
29 sort(rd+1,rd+m+1,cmp);
30 dp[0][0]=1;
31 for(int i=1,now=1,pre=0;i<=m;i++)
32 {
33 //printf("%d\n",rd[i]);
34 for(int j=0;j<=k;j++)
35 if(dp[pre][j]&&j+cnt[rd[i]]<=k&&n-j-cnt[rd[i]]<=k)
36 ans=min(ans,1LL*(j+cnt[rd[i]])*ls[rd[i]]+1LL*(n-j-cnt[rd[i]])*n);
37 for(int j=0;j<=k;j++)
38 {
39 dp[now][j]|=dp[pre][j];
40 if(j+cnt[rd[i]]<=k)
41 dp[now][j+cnt[rd[i]]]|=dp[pre][j];
42 dp[pre][j]=0;
43 }
44 swap(now,pre);
45 }
46 if(ans==2e18) printf("-1\n");
47 else printf("%lld\n",ans);
48 return 0;
49 }