ZOJ 1877 Bridge（贪心）

不要乱猜题目，要分情况讨论，通过验证得出结论，这就是贪心。

题目的主要意思是一群人要过桥，每次至多两个人一起过，过桥需要手电筒，而他们只有一个手电筒，这意味着每次都要有一个人把手电筒送回来。每个人的速度不同，过桥用的时间也不一样，求最短的时间。

     主要先理解算法思路：先分析四个人过河的情况，然后再推广到 n 个人。

   1. 先讨论四个人的情况，一共是有两种方案的：
       假设四个步速为a < b < c < d, 
             方案一：AB过A回 CD过B回 AB过，时间为 b + a + d + b + b；
             方案二：AC过A回 AD过A回 AB过 ，时间为 c + a + d + a + b；
      比较两者时间，关键在 2*b 与 a+c 的大小 。

    2. 当人数大于四时，四步一组，每次优先将最慢 的两个送到对岸；当人数小于四时，单独计算。

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 ashlight. Only one ashlight is available among the

n

people, so some sort of

shuttle arrangement must be arranged in order to return the ashlight 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 input begins with a single positive integer on a line by itself indicating the number of the cases

following, each of them as described below. This line is followed by a blank line, and there is also a

blank line between two consecutive inputs.

The rst 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.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases

will be separated by a blank line.

The rst 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 speci ed 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 ashlight so that more

may cross. If more than one strategy yields the minimal time, any one will do.

SampleInput

1

4

1

2

5

10

SampleOutput

17

1 2

1

5 10

2

1 2

AC代码：

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>#include<cstring>#include<string>#include<vector>#include<queue>#include<map>#include<set>#define INF 999999999#define LL long long#define mod 1000003using namespace std;int main(){int t;scanf("%d",&t);while(t--)    {        int n;        scanf("%d",&n);        int a[1005];        for(int i=0;i<n;i++)            scanf("%d",a+i);        int sum=0;        sort(a,a+n);        int cnt=0;        int j=n-1;        while(cnt<=n-4)        {            if(2*a[1]<a[0]+a[j-1])                sum+=a[1]+a[0]+a[j]+a[1];            else                sum+=a[j-1]+a[0]+a[j]+a[0];            cnt+=2;            j-=2;        }        switch(n-cnt)        {            case 1:sum+=a[0];break;            case 2:sum+=a[1];break;            case 3:sum+=a[0]+a[1]+a[2];break;        }        printf("%d\n",sum);        cnt=0;        j=n-1;        while(cnt<=n-4)        {            if(2*a[1]<a[0]+a[j-1])                {                    printf("%d %d\n%d\n",a[0],a[1],a[0]);                    printf("%d %d\n%d\n",a[j-1],a[j],a[1]);                }            else                {                    printf("%d %d\n%d\n",a[0],a[j-1],a[0]);                    printf("%d %d\n%d\n",a[0],a[j],a[0]);                }            cnt+=2;            j-=2;        }        switch(n-cnt)        {            case 1:printf("%d\n",a[0]);break;            case 2:printf("%d %d\n",a[0],a[1]);break;            case 3:printf("%d %d\n%d\n%d %d\n",a[0],a[1],a[0],a[0],a[2]);break;        }        if(t!=0)            printf("\n");    }}