zoukankan      html  css  js  c++  java
  • 2017 ICPC亚洲区域赛北京站 J Pangu and Stones(区间dp)

    In Chinese mythology, Pangu is the first living being and the creator of the sky and the earth. He woke up from an egg and split the egg into two parts: the sky and the earth.

    At the beginning, there was no mountain on the earth, only stones all over the land.

    There were N piles of stones, numbered from 1 to N. Pangu wanted to merge all of them into one pile to build a great mountain. If the sum of stones of some piles was S, Pangu would need S seconds to pile them into one pile, and there would be S stones in the new pile.

    Unfortunately, every time Pangu could only merge successive piles into one pile. And the number of piles he merged shouldn't be less than L or greater than R.

    Pangu wanted to finish this as soon as possible.

    Can you help him? If there was no solution, you should answer '0'.

    Input

    There are multiple test cases.

    The first line of each case contains three integers N,L,R as above mentioned (2<=N<=100,2<=L<=R<=N).

    The second line of each case contains N integers a1,a2 …aN (1<= ai  <=1000,i= 1…N ), indicating the number of stones of  pile 1, pile 2 …pile N.

    The number of test cases is less than 110 and there are at most 5 test cases in which N >= 50.

    Output

    For each test case, you should output the minimum time(in seconds) Pangu had to take . If it was impossible for Pangu to do his job, you should output  0.

    Sample Input

    3 2 2
    1 2 3
    3 2 3
    1 2 3
    4 3 3
    1 2 3 4

    Sample Output

    9
    6
    0

    题意:

    n个石子堆排成一排,每次可以将连续的[L,R]堆石子合并成一堆,花费为要合并的石子总数。求将所有石子合并成一堆的最小花费,如无法实现则输出0。

    思路:

    dp[i][j][k]表示将区间[i, j]合并成k堆的最小代价,转移有:

    k=1时:

    dp[i][j][1]=min(dp[i][j][1],dp[i][j][q]+sum[j]-sum[i-1]) 

    k>1时:

    dp[i][j][q]=min(dp[i][j][q],dp[i][k][q-1]+dp[k+1][j][1]) 

    #include<bits/stdc++.h> 
    using namespace std;
    #define MAX 105
    #define INF 0x3f3f3f3f
    int sum[MAX],dp[MAX][MAX][MAX];
    int main()
    {
        int n,l,r,i,j,k;
        while(scanf("%d%d%d",&n,&l,&r)!=EOF)
        {
            memset(dp,INF,sizeof(dp));
            for(i=1;i<=n;i++)
            {
                scanf("%d",&sum[i]);
                dp[i][i][1]=0;
                sum[i]+=sum[i-1];
            }
            int len;
            for(len=l;len<=r;len++)    //merge长度 len[l,r] 
            {
                for(i=1;i+len-1<=n;i++)//merge范围 [i,i+len-1]
                {
                    j=i+len-1;
                    dp[i][j][len]=0;
                    dp[i][j][1]=sum[j]-sum[i-1];
                }
            }
        
            int q;
            for(len=2;len<=n;len++)    //merge长度 len[2,n] 
            {
                for(i=1;i+len-1<=n;i++)//merge范围 [i,i+len-1]
                {
                    j=i+len-1;
                    for(k=i;k<j;k++)
                        for(q=2;q<=len;q++)
                            dp[i][j][q]=min(dp[i][j][q],dp[i][k][q-1]+dp[k+1][j][1]);
                    for(q=l;q<=r;q++)
                        dp[i][j][1]=min(dp[i][j][1],dp[i][j][q]+sum[j]-sum[i-1]);
                } 
            }
        
            if(dp[1][n][1]<INF)
                printf("%d
    ",dp[1][n][1]);
            else printf("0
    ");
        }
        return 0;
    }
  • 相关阅读:
    机器学习概要
    Latex公式压缩
    MATLAB多项式运算
    利用MathType为公式编号并引用
    MATLAB符号对象与符号运算
    MATLAB矩阵运算
    MATLAB绘制函数图
    MATLAB程序控制语句
    MATLAB关系运算符和逻辑运算符
    Raspberry pi之wifi设置-3
  • 原文地址:https://www.cnblogs.com/kannyi/p/9962443.html
Copyright © 2011-2022 走看看