【动态规划】石子合并 luogu-1880

分析

简单的区间DP

AC代码

#include <bits/stdc++.h>
using namespace std;
#define ms(a,b) memset(a,b,sizeof(a))
typedef long long ll;
int Dp_max[305][305],Dp_min[305][305];
int a[305],sum[305];
inline int read(){
    int X=0,w=0; char ch=0;
    while(!isdigit(ch)) {w|=ch=='-';ch=getchar();}
    while(isdigit(ch)) X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
    return w?-X:X;
}
int main(int argc,char* argv[]){
	int n=read();
	for (int i=1;i<=n;i++) a[i]=read(),a[i+n]=a[i];
	for (int i=1;i<=(n<<1);i++) sum[i]=sum[i-1]+a[i];
	for (int i=(n<<1);i>=1;i--) {
		for (int j=i+1;j<i+n;j++) {
			Dp_min[i][j]=1<<30;
			for (int k=i;k<j;k++) {
				Dp_max[i][j]=max(Dp_max[i][j],Dp_max[i][k]+Dp_max[k+1][j]+sum[j]-sum[i-1]);
				Dp_min[i][j]=min(Dp_min[i][j],Dp_min[i][k]+Dp_min[k+1][j]+sum[j]-sum[i-1]);
			}
		}
	}
	int ans1=1<<30,ans2=0;
	for (int i=1;i<=n;i++) {
		ans1=min(ans1,Dp_min[i][i+n-1]);
		ans2=max(ans2,Dp_max[i][i+n-1]);
	}
	printf("%d
%d
",ans1,ans2);
	return 0;
}