bzoj 1863: [Zjoi2006]trouble 皇帝的烦恼【二分+dp】

二分答案，注意l是max(a[i]+a[i+1])，r是sum_a
判断的时候用dp，设f[i]为i与1最少的相同颜色数，g[i]为i与1最多的相同颜色数，转移是f[i]=max(a[i]-(w-a[i-1]-(a[1]-g[i-1])),0),g[i]=min(a[1]-f[i-1],a[i]);

#include<iostream>
#include<cstdio>
using namespace std;
const int N=20005;
int n,a[N],f[N],g[N];
int read()
{
	int r=0,f=1;
	char p=getchar();
	while(p>'9'||p<'0')
	{
		if(p=='-')
			f=-1;
		p=getchar();
	}
	while(p>='0'&&p<='9')
	{
		r=r*10+p-48;
		p=getchar();
	}
	return r*f;
}
bool ok(int w)
{
	f[1]=a[1],g[1]=a[1];
	for(int i=2;i<=n;i++)
		f[i]=max(a[i]-(w-a[i-1]-(a[1]-g[i-1])),0),g[i]=min(a[1]-f[i-1],a[i]);
	return f[n]==0;
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
		a[i]=read();
	a[n+1]=a[1];
	int l=0,r=0,ans=0;
	for(int i=1;i<=n;i++)
		l=max(l,a[i]+a[i+1]),r+=a[i];
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(ok(mid))
			r=mid-1,ans=mid;
		else
			l=mid+1;
	}
	printf("%d\n",ans);
	return 0;
}