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;
}