1048 - Conquering Keokradong

Posted 2020-07-23 SJY

1048 - Conquering Keokradong

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Time Limit: 1 second(s)

Memory Limit: 32 MB

This winter we are going on a trip to Bandorban. The main target is to climb up to the top of Keokradong. So, we will use a trail. The trail is a continuous marked footpath that goes from Bandorban to Keokradong.

Part of the experience is also the route planning of the trip. We have a list of all possible campsites that we can use along the way and we want to do this trip so that we only stop K nights to camp. We also know in advance the distance between consecutive campsites and we are only allowed to camp at a campsite. Our goal is to plan the trip so that we minimize the maximum amount of walking done in a single day. In other words, if our trip involves 2 nights (3 days of walking), and we walk 9, 10, 5 miles on each day respectively, the cost (maximum amount of walking done in one day) is 10. Another schedule that involves walking 9, 6, 9 miles on each day has cost 9.

Given the distances between N consecutive campsites of a trail and given the number of nights for your trip, K, your task is to devise a camping strategy for the specified trail such that it minimizes the maximum amount of walking done in a single day. Note that the first distance value given is the distance from our start-point of the trail to our 1^st campsite, and the last distance value given is the distance from our N^th campsite to our end-point of the trail.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains of two integers, the number of campsites, N (1 ≤ N ≤ 1000) and the number of nights of the trip, K (1 ≤ K ≤ min(N, 300)). The following N + 1 lines indicate the distance in miles between consecutive campsite locations. All the integers will be positive and less than 10000.

Output

For each case of input you have to print the case number and the minimized cost as described above. Then print K+1 lines, each containing the amount of distance covered in i^th day. As there can be many solutions, the primary target is to find the one which ensures that each day we have to walk some distance. For ties, print the one where the distance covered in first day is maximum, then the distance covered in second day is maximum and so on.

Sample Input	Output for Sample Input
1 4 3 7 2 6 4 5	Case 1: 8 7 8 4 5

 

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PROBLEM SETTER: JANE ALAM JAN

题意：将N+1个数分成K+1段，并且求这些段中的最大值的最小是多少，并且保证最小的情况下按照第一段最大优先，然后第二段。。。。

思路：二分+贪心

先用二分去找最大值的最小是多少，我们可以知道当我们分的段数越小那么这个最大的值就越大，所以我们二分找到，可以分成<=k+1段的最小的最大值

然后我们知道这个值的时候，贪心组合按前到后，贪心选到每段的最大，最后只要保证取到K+1段就行

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<queue>
 6 #include<stack>
 7 #include<set>
 8 #include<math.h>
 9 using namespace std;
10 int ans[2000];
11 int uu[2000];
12 bool check(int k,int n,int m)
13 {
14         int i,j;
15         int sum=0;
16         int cnt=1;
17         for(i=0; i<=n; i++)
18         {
19                         if(sum+ans[i]>k)
20                         {
21                                 uu[cnt-1]=sum;
22                                 sum=ans[i];
23                                 cnt++;
24                         }
25                         else if(sum+ans[i]<=k)
26                         {
27                                 sum+=ans[i];
28                         }
29         }uu[cnt-1]=sum;
30         if(m>=cnt)
31                 return true;
32         else return false;
33 }
34 int main(void)
35 {
36         int i,j,k;
37         int s;
38         scanf("%d",&k);
39         for(s=1; s<=k; s++)
40         {       memset(uu,0,sizeof(uu));
41                 int n;
42                 int m;
43                 int maxx=0;
44                 int  sum=0;
45                 scanf("%d %d",&n,&m);
46                 for(i=0; i<=n; i++)
47                 {
48                         scanf("%d",&ans[i]);
49                         maxx=max(maxx,ans[i]);
50                         sum+=ans[i];
51                 }
52                 int l=maxx;
53                 int r=sum;
54                 int answer=-1;
55                 while(l<=r)
56                 {
57                         int mid=(l+r)/2;
58                         bool us=check(mid,n,m+1);
59                         if(us)
60                         {
61                                 answer=mid;
62                                 r=mid-1;
63                         }
64                         else l=mid+1;
65                 }
66                 printf("Case %d:",s);
67                 printf(" %d\n",answer);
68                 check(answer,n,m);
69                 int ac=0; sum=0;
70                 int cnt=1;
71                for(i=0;i<=n;i++)
72                {
73                    if(sum+ans[i]>answer||(n-i-1)<m-cnt)
74                    {
75                        uu[cnt-1]=sum;
76                        sum=ans[i];
77                        cnt++;
78                    }
79                    else
80                    {
81                        sum+=ans[i];
82                    }
83                }
84                uu[cnt-1]=sum;
85                for(i=0;i<m+1;i++)
86                {
87                    printf("%d\n",uu[i]);
88                }
89         }
90         return 0;
91 }