2017ICPC北京 J：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：

n个石子堆排成一排，每次可以将连续的最少L堆，最多R堆石子合并在一起，消耗的代价为要合并的石子总数

求合并成1堆的最小代价，如果无法做到输出0

思路：

dp[i][j][k]表示区间[i, j]分成k堆的最小代价，转移有

k=1时：

dp[i][j][1] = min(dp[i][p][x-1]+dp[p+1][j][1]+sum[i][j] (i<=p<=j-1；L<=x<=R) )

k>1时：

dp[i][j][k] = min(dp[i][p][k-1]+dp[p+1][j][1] ( i<=p<=j-1) )

// 2017ICPC北京
#include
#include
#include
using namespace std;
#define LL long long
int a[105], sum[105][105];
LL dp[105][105][105];
int main(void)
{
	int i, j, k, p, n, l, r;
	while(scanf("%d%d%d", &n, &l, &r)!=EOF)
	{
		for(i=1;i<=n;i++)
			scanf("%d", &a[i]);
		memset(dp, 62, sizeof(dp));
		for(i=1;i<=n;i++)
		{
			sum[i][i-1] = 0;
			for(j=i;j<=n;j++)
			{
				sum[i][j] = sum[i][j-1]+a[j];
				dp[i][j][j-i+1] = 0;
			}
		}
		for(i=1;i<=n;i++)
		{
			for(j=i;j<=n;j++)
				dp[i][j][j-i+1] = 0;
		}
		for(p=1;p<=n-1;p++)
		{
			for(i=1;i+p<=n;i++)
			{
				for(j=i;j<=i+p-1;j++)
				{
					for(k=l-1;k<=r-1;k++)
						dp[i][i+p][1] = min(dp[i][i+p][1], dp[i][j][k]+dp[j+1][i+p][1]+sum[i][i+p]);
				}
				for(j=2;j<=p;j++)
				{
					for(k=i;k<=i+p-1;k++)
						dp[i][i+p][j] = min(dp[i][i+p][j], dp[i][k][j-1]+dp[k+1][i+p][1]);
				}
			}
		}
		if(dp[1][n][1]==4485090715960753726ll)
			printf("0\n");
		else
			printf("%lld\n", dp[1][n][1]);
	}
	return 0;
}
/*
2 1 1
1 2
*/