Garlands - UVa 1443 二分+dp

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本文介绍了一道关于圣诞灯饰布置的算法题目，旨在寻找最优的灯串分割方式，以确保每半段的重量和达到最小的最大值。文章给出了问题背景、输入输出格式、样例及解析，并提供了AC代码实现。

Garland decorator is a profession which recently gained in importance, especially during Christmas time. Any kid can decorate a christmas tree, any parent can put gifts in sockets, and even anyone can start believing in Santa Claus, but hanging christmas garlands is a completely difflerent story. As you will learn, it is an extremely important, responsible and tough job.

A garland consists of n pieces of equal length. Due to decorations like christmas balls attached to garlands, piece i has its own weight w_i. The garland has to be attached to the ceiling in m spots, where the very beginning of the garland should be attached to spot 1 and its end to spot m. The garland should also be hooked to the remaining spots, which divides it into segments, each consisting of several consecutive pieces. There are, however, several rules that every respectable garland decorator should keep in mind.

(i)

Each segment should contain a positive even number of pieces. Due to this condition, we may divide a segment into two half-segments.

(ii)

To minimize the chance that a guest hits your precious garland with their head (and tears it into pieces), the garland cannot hang too low: each half-segment can contain at most d pieces.

(iii)

Finally, to keep the ceiling from falling on people heads, the decorator should minimize the weight of the heaviest half-segment.

An example of an optimally hanging garland (consisting of twelve pieces in three segments) is presented below; weights of respective pieces are given in circles.

$\epsfbox{p4625.eps}$

Input

The input contains several test cases. The first line of the input contains a positive integer Z $\le$ 50, denoting the number of test cases. Then Z test cases follow, each conforming to the format described below.

The description of each garland consists of two lines. The first line describing a particular garland contains three positive integers n, m, and d (1 $\le$ n $\le$ 40000, 2 $\le$ m $\le$ 10000, 1 $\le$ d $\le$ 10000) separated by single spaces and described above. The second line contains n positive integers w₁, w₂,..., w_n (1 $\le$ w_i $\le$ 10000), being the weights of the corresponding pieces.

Output

For each test case, your program has to write an output conforming to the format described below.

For each garland, your program should output a single line containing one integer, being the weight of the heaviest half-segment in an optimal attachment of the garland. If it is not possible to hang the garland satisfying conditions (i) and (ii), then your program should output word `BAD'.

Sample Input

4 
4 3 10 
10 10 20 20 
6 4 10 
1 1 100 100 1 1 
6 3 10 
1 1 100 100 1 1 
1 2 2 
333

Sample Output

题意：将一个长度为n的灯分成m-1段，每段必须是偶数个，且每半段的数量不超过d，问每半段的重量和的最大值的最小值。

思路：先排除不可能的情况，然后二分这个重量的答案，用dp进行验证，dp[i]表示前i的灯至少需要分多少段，同时dp[i][0,1]区分奇数和偶数。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
typedef long long ll;
ll val[40010],sum[40010];
int n,n2,m,d,dp[20010][2],INF=1e9;
bool check(ll maxn)
{
    int i,j;
    dp[0][0]=0;
    dp[0][1]=INF;
    for(i=1;i<=n2;i++)
    {
        dp[i][0]=INF;
        dp[i][1]=INF;
        for(j=i-1;j>=0 && i-j<=d;j--)
        {
            if(sum[i*2]-sum[i+j]>maxn)
             break;
            if(sum[i+j]-sum[j*2]<=maxn)
            {
                dp[i][0]=min(dp[i][0],dp[j][1]+1);
                dp[i][1]=min(dp[i][1],dp[j][0]+1);
            }
        }
    }
    if(dp[n2][m%2]<=m)
      return true;
    else
      return false;
}
int main()
{
    int T,t,i,j,k,a,b,len;
    ll l,r,mi;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%d%d%d",&n,&m,&d);
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&val[i]);
            sum[i]=sum[i-1]+val[i];
        }
        if(m==1 || n&1 || n<2*(m-1) || n>(m-1)*d*2)
        {
            printf("BAD\n");
            continue;
        }
        n2=n/2;
        m--;
        l=1;r=sum[n];
        while(l<r)
        {
            mi=(l+r)/2;
            if(check(mi))
              r=mi;
            else
              l=mi+1;
        }
        printf("%lld\n",l);
    }
}