hihocoder 1636 : Pangu and Stones(区间dp)

1636 : Pangu and Stones

时间限制:1000ms
单点时限:1000ms
内存限制:256MB
描述
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’.

输入
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.

输出
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.

样例输入
3 2 2
1 2 3
3 2 3
1 2 3
4 3 3
1 2 3 4
样例输出
9
6
0

题意：给出n堆石头，每次最少合并其中l堆，最多合并r堆，问合成1堆最少需要花费多少时间

dp[i][j][k]表示i~j这个区间合成k堆所需要的最小时间，故可得状态转移方程式：
d为枚举的区间间隔
1.k==1 dp[i][i+d][1]=min(dp[i][i+d][1],dp[i][j][k]+dp[j+1][i+d][1]+sum[i][i+d])
(l-1<=k<=r-1)
2.k>=2 dp[i][i+d][k]=min(dp[i][i+d][k],dp[i][j][k-1]+dp[j+1][i+d][1])
此处k不用做限制

写这题的时候一直在考虑如何限制每次合并的堆数在l~r这个范围内，最后也没搞出来，事实上，只需要在合并一堆的时候限制条件就行了，因为所有k>2的情况都是由k=1的情况得出的，所以在都初始化为inf的情况下，不能合成1堆，dp[i][j][1]=inf,那么后面所有由dp[i][j][1]推出的情况也是inf

#include
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int N=105;
const int inf=0x3f3f3f3f;
int n,l,r;
int w[N];
int sum[N][N];
int dp[N][N][N];//i~j区间分成k堆最小价格
int main()
{
    while(~scanf("%d%d%d",&n,&l,&r))
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&w[i]);
        mem(dp,inf);
        for(int i=1; i<=n; i++)
        {
            sum[i][i-1]=0;
            for(int j=i; j<=n; j++)
            {
                sum[i][j]=sum[i][j-1]+w[j];
                dp[i][j][j-i+1]=0;//初始化初状态
            }
        }
        for(int d=1; d<=n; d++)
            for(int i=1; i+d<=n; i++)
            {
                for(int j=i; j<=i+d-1; j++)
                    for(int k=l-1; k<=r-1; k++)
                        {
                            dp[i][i+d][1]=min(dp[i][i+d][1],dp[i][j][k]+dp[j+1][i+d][1]+sum[i][i+d]);
                        }
                for(int k=2; k<=d; k++)
                    for(int j=i; j<=i+d-1; j++)
                        dp[i][i+d][k]=min(dp[i][i+d][k],dp[i][j][k-1]+dp[j+1][i+d][1]);
            }
        if(dp[1][n][1]==inf)
            puts("0");
        else
            printf("%d\n",dp[1][n][1]);
    }
    return 0;
}