CF82D Two out of Three

题目描述

Vasya has recently developed a new algorithm to optimize the reception of customer flow and he considered the following problem.

Let the queue to the cashier contain n n n people, at that each of them is characterized by a positive integer ai a_{i} ai​ — that is the time needed to work with this customer. What is special about this very cashier is that it can serve two customers simultaneously. However, if two customers need ai a_{i} ai​ and aj a_{j} aj​ of time to be served, the time needed to work with both of them customers is equal to max(ai,aj) max(a_{i},a_{j}) max(ai​,aj​) . Please note that working with customers is an uninterruptable process, and therefore, if two people simultaneously come to the cashier, it means that they begin to be served simultaneously, and will both finish simultaneously (it is possible that one of them will have to wait).

Vasya used in his algorithm an ingenious heuristic — as long as the queue has more than one person waiting, then some two people of the first three standing in front of the queue are sent simultaneously. If the queue has only one customer number i i i , then he goes to the cashier, and is served within ai a_{i} ai​ of time. Note that the total number of phases of serving a customer will always be equal to ⌈n/2⌉ ⌈n/2⌉ ⌈n/2⌉ .

Vasya thinks that this method will help to cope with the queues we all hate. That's why he asked you to work out a program that will determine the minimum time during which the whole queue will be served using this algorithm.

输入格式

The first line of the input file contains a single number n n n ( 1<=n<=1000 1<=n<=1000 1<=n<=1000 ), which is the number of people in the sequence. The second line contains space-separated integers a1,a2,...,an a_{1},a_{2},...,a_{n} a1​,a2​,...,an​ ( 1<=ai<=106 1<=a_{i}<=10^{6} 1<=ai​<=106 ). The people are numbered starting from the cashier to the end of the queue.

输出格式

Print on the first line a single number — the minimum time needed to process all n n n people. Then on ⌈n/2⌉ ⌈n/2⌉ ⌈n/2⌉ lines print the order in which customers will be served. Each line (probably, except for the last one) must contain two numbers separated by a space — the numbers of customers who will be served at the current stage of processing. If n n n is odd, then the last line must contain a single number — the number of the last served customer in the queue. The customers are numbered starting from 1 1 1 .

题意翻译

一队顾客排在一位收银员前面。他采取这样一个策略：每次，假如队伍有至少两人，就会从前面的前三人（如果有）中选取两位一起收银，所花费的时间为这两人单独收银所需时间的最大值。如果只有两人，那么一起收银；如果只有一人，那么单独收银。请问所需的总时间最少是多少？

输入输出样例

输入 #1

4
1 2 3 4

输出 #1

6
1 2
3 4

输入 #2

5
2 4 3 1 4

输出 #2

8
1 3
2 5
4

考虑dp

发现不好搞，在不同的情况下选择不同的数会造成不同的影响
考虑状态的设计，发现对于不同的情况，不一样的其实是当前3个数和选到了第几个数
可以在状态中记录一下目前3个数，发现只用记录前面留下的即可

设f[i][j]表示选到了第i个数，前面留下的是第j个数时的最优解
然后就没什么了
转移应该是很显而易见的

code

//¼ÓÓÍ
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int f[1001][1001],n,a[1001],out[1001][1001][3];
inline ll read()
{
    char c=getchar();ll a=0,b=1;
    for(;c<'0'||c>'9';c=getchar())if(c=='-')b=-1;
    for(;c>='0'&&c<='9';c=getchar())a=a*10+c-48;
    return a*b;
}
inline void az(int i,int j,int x,int y,int z){out[i][j][0]=x;out[i][j][1]=y;out[i][j][2]=z;}
void prout(int x,int y)
{
    if(y==0)return;
    prout(x-1,out[x][y][2]);
    if(out[x][y][0]<=n)
    {
        cout<<out[x][y][0]<<' ';
    }
    if(out[x][y][1]<=n)
    {
        cout<<out[x][y][1]<<' ';
    }
    cout<<endl;
}
int main()
{
//    freopen(".in","r",stdin);
//    freopen(".out","w",stdout);
    n=read();
    for(int i=1;i<=n;i++)
    {
        a[i]=read();
    }
    memset(f,0x3f,sizeof(f));
    f[1][1]=max(a[2],a[3]);f[1][2]=max(a[1],a[3]);f[1][3]=max(a[1],a[2]);
    az(1,1,2,3,0);
    az(1,2,1,3,0);
    az(1,3,1,2,0);
    int m=(n&1)?n/2+1:n/2;
    for(int i=2;i<=m;i++)
    {
        int x=i<<1,y=i<<1|1;
        for(int j=1;j<x;j++)
        {
            if(f[i][x]>f[i-1][j]+max(a[j],a[y]))
            {
                f[i][x]=f[i-1][j]+max(a[j],a[y]);
                az(i,x,j,y,j);
            }
            if(f[i][y]>f[i-1][j]+max(a[j],a[x]))
            {
                f[i][y]=f[i-1][j]+max(a[j],a[x]);
                az(i,y,j,x,j);
            }
            if(f[i][j]>f[i-1][j]+max(a[x],a[y]))
            {
                f[i][j]=f[i-1][j]+max(a[x],a[y]);
                az(i,j,x,y,j);
            }
        }
    }
    cout<<f[m][n+1]<<endl;
    prout(m,n+1);
    return 0;
}