[Problem solution] Luogu P1880 [NOI1995] Stone merger

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topic

Luogu P1880 stone merger

answer

This is a classic interval dp question, f[l][r] represents the maximum value after combining the stones in the interval [l,r]. The information of the long-length section can be derived from the small-length section. The data in the question is circular, you need to break the loop into a chain, copy the original sequence and add it to the back.

Code

#include <bits/stdc++.h>
using namespace std;
const int inf=0x7fffffff;
const int maxn=200+5;
int n;
int f1[maxn][maxn],f2[maxn][maxn],a[maxn],sum[maxn];
//sum[]是前缀和,f1,f2[][]分别表示最小/大值 
inline void dp()
{
	for(int i=1;i<=(n<<1);i++)for(int j=i+1;j<=(n<<1);j++) f1[i][j]=inf,f2[i][j]=-inf; 
	for(int len=2;len<=n;len++) //区间长度 
	{
		for(int l=1; ;l++) //区间左端点 
		{
			int r=l+len-1; if(r>(n<<1)) break;
			for(int k=l;k<r;k++) //枚举断点 
			{
				f1[l][r]=min(f1[l][r],f1[l][k]+f1[k+1][r]+sum[r]-sum[l-1]);
				f2[l][r]=max(f2[l][r],f2[l][k]+f2[k+1][r]+sum[r]-sum[l-1]);
			}
		}
	}
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {scanf("%d",&a[i]); a[i+n]=a[i];}  
	for(int i=1;i<=(n<<1);i++) sum[i]=sum[i-1]+a[i];
	dp(); int ans_min=inf,ans_max=-inf;
	for(int l=1; ;l++) 
	{
		if(l>n+1) break;
		ans_min=min(ans_min,f1[l][l+n-1]);
		ans_max=max(ans_max,f2[l][l+n-1]);
	}
	printf("%d\n%d\n",ans_min,ans_max);
	
	return 0;
}