ICPC2017Beijing J Pangu and Stones(铜牌题)

Problem

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.

Solution

石子合并，每次合并相邻的一些堆，代价是合并的石子数之和，每次合并的堆数有上下限，求最小的代价。
dp[i][j][k]表示从i合并到j，划分成k堆的最小代价。
枚举要计算的区间长度，枚举具体的区间，枚举划分的堆数，枚举具体划分方法。
dp[i][j][k]=min(dp[i][j][k],dp[i][c][k-1]+dp[c+1][j][1]);
然后枚举符合条件的划分堆和具体方法，合并成一堆。
dp[i][j][1]=min(dp[i][j][1],dp[i][c][k]+dp[c+1][j][1]+sum[j]-sum[i-1]);

Code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<cstring>
#define ll long long
#define ioss ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;
int dp[N][N][N], s[N], sum[N];
int n,l,r;
int main(){
    ioss;
    //freopen("51nod_1880_5_in.txt","r",stdin);
    //freopen("ans.out","w",stdout);
    while(cin>>n>>l>>r){
        memset(dp,0x3f,sizeof(dp));
        for(int i=1;i<=n;i++){
            cin>>s[i];
            sum[i]=sum[i-1]+s[i];
            dp[i][i][1]=0;
        }
        for(int len=2;len<=n;len++){
            for(int i=1;i+len-1<=n;i++){
                int j=i+len-1;
                for(int k=2;k<=min(len,r);k++){
                    for(int c=i+k-2;c<j;c++){
                        dp[i][j][k]=min(dp[i][j][k],dp[i][c][k-1]+dp[c+1][j][1]);
                    }
                }
                for(int k=l-1;k<=r-1;k++){
                    for(int c=i+k-2;c<j;c++){
                        dp[i][j][1]=min(dp[i][j][1],dp[i][c][k]+dp[c+1][j][1]+sum[j]-sum[i-1]);
                    }
                }
            }
        }
        cout<<(dp[1][n][1]==INF?0:dp[1][n][1])<<endl;
    }
    return 0;
}