ZOJ Problem Set - 1877

Bridge

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Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

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n people wish to cross a bridge at night. A group of at most two people may cross at any time, and each group must have a flashlight. Only one flashlight is available among the n people, so some sort of shuttle arrangement must be arranged in order to return the flashlight so that more people may cross.

Each person has a different crossing speed; the speed of a group is determined by the speed of the slower member. Your job is to determine a strategy that gets all n people across the bridge in the minimum time.

Input

The first line of input contains n, followed by n lines giving the crossing times for each of the people. There are not more than 1000 people and nobody takes more than 100 seconds to cross the bridge.

Input contains multiple test cases. Process to the end of file.

Output

The first line of output must contain the total number of seconds required for all n people to cross the bridge. The following lines give a strategy for achieving this time. Each line contains either one or two integers, indicating which person or people form the next group to cross. (Each person is indicated by the crossing time specified in the input. Although many people may have the same crossing time the ambiguity is of no consequence.) Note that the crossings alternate directions, as it is necessary to return the flashlight so that more may cross. If more than one strategy yields the minimal time, any one will do.

Sample Input

4
1
2
5
10

Sample Output

17
1 2
1
5 10
2
1 2

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Source: University of Waterloo Local Contest 2000.09.

題目分析：

　　有n我的經過一座橋到橋的另外一頭，每次只能經過兩我的而且須要一把手電，不一樣的人過橋的速度不一樣，每次經過的時間由速度最慢的那我的決定，過橋後必須有一我的把手電帶回去。

求全部人都到橋的另一頭所需的最少時間。

輸出最少時間和每一次過橋的人。

 

由於每次過橋後，都有一我的會回來，那麼使用貪心的思想，每次都由速度最快的人帶一個慢的人過橋，而後快的人再回來。

設花費的時間爲time；

考慮只有a一我的的狀況：

　　直接過橋。

　　time=a;

考慮只有a,b兩我的的狀況：

　　直接過橋。

　　time=max(a,b);

考慮只有a,b,c三我的的狀況:

　　假設a<b<c.

　　a b 過橋，a回來，a c過橋

　　time=b+a+c;

考慮只有a,b,c,d 4我的的狀況：

　　假設a<b<c<d.

　　使用貪心思想，會有兩種狀況：

　　1)、a b過橋，a回來，c d過橋，b回來，a b過橋；

　　time=b+a+d+b+b;

　　2）、a c過橋，a回來，b d過橋 ，b回來，a b過橋；

　　time=c+a+d+b+b;

　　結果去兩種狀況的最小值；

　　

當n>4時:

　　S={a,b,c,d,e,f....};

將全部人按照速度排序，最慢的人爲j，只需考慮S0,S1,Sj-1,Sj的過橋狀況。這樣不斷的將最慢的人過橋，就把n的狀況轉換爲前面的三種平凡的狀況。

 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <vector>
 5 #include <algorithm>
 6 #include <cstring>
 7 #include <map>
 8 using namespace std;
 9 const int N=1009;
10 int s[N];
11 int n;
12 
13 int main()
14 {
15     int time=0;
16     int count=0;
17     int j;
18     while(scanf("%d",&n)!=EOF)
19     {
20         time=0;
21         count =0;
22 
23         for(int i=0;i<n;i++)
24         {
25             scanf("%d",&s[i]);
26         }
27         sort(s,s+n);
28         j=n-1;//最慢的人
29         while(count<=n-4)//4我的的狀況
30         {
31             if(2*s[1]<s[0]+s[j-1])
32             {
33                 time+=s[1]+s[1]+s[j]+s[0];
34             }
35             else
36             {
37                 time+=s[j]+s[0]+s[0]+s[j-1];
38             }
39             count+=2;//有2我的過橋
40             j-=2;//規模減小2
41         }
　　　　　　//剩下的3種狀況
42         switch(n-count)
43         {
44             case 1:
45             time+=s[0];
46             break;
47             case 2:
48             time+=s[1];
49             break;
50             case 3:
51             time+=s[0]+s[1]+s[2];
52             break;
53         }
54         printf("%d\n",time);//輸出總時間
55         j=n-1;
56         count =0;
　　　　　　//輸出過橋的方法
57         while(count<=n-4)
58         {
59             if(2*s[1]<s[0]+s[j-1])
60             {
61                 printf("%d %d\n%d\n",s[0],s[1],s[0]);
62                 printf("%d %d\n%d\n",s[j-1],s[j],s[1]);
63             }
64             else
65             {
66                 printf("%d %d\n%d\n",s[0],s[j-1],s[0]);
67                 printf("%d %d\n%d\n",s[0],s[j],s[0]);
68             }
69             count+=2;
70             j-=2;
71             
72         }
73         switch(n-count)
74                 {
75             case 1:
76             printf("%d\n",s[0]);
77             break;
78             case 2:
79             printf("%d %d\n",s[0],s[1]);
80             break;
81             case 3:
82             printf("%d %d\n%d\n",s[0],s[1],s[0]);
83             printf("%d %d\n",s[0],s[2]);
84             break;
85                 }
86     }
87     return 0;
88 }