luogu 2115 破坏(01分数规划)

题意:给出一个序列，删除一个连续的子串后使得剩下的平均值最小。

 

典型的01分数规划，令f(x)=(sum1[i]+sum2[j])/(i+j).sum1表示前缀和，sum2表示后缀和，那么我们就相当于求出f(x)的最小值。

令f(x)=y,化简则有(sum1[i]-i*y)+(sum2[j]-j*y)=0,我们二分y，找出满足这个式子的y的最小值。

根据这个式子可以把序列都减去一个y，这样就相当于求新序列的前缀和sum1[i]+sum2[j]>=0.

实际上就是求min(sum1[i]+sum2[j])>=0,转化一下就变成了求新序列的最大子串和的经典DP问题。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-7
# define MOD 1024523
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=100005;
//Code begin...

int a[N], n;
double b[N], dp[N], sum;

bool check(double x){
   sum=0;
   FOR(i,1,n) b[i]=a[i]-x, sum+=b[i];
   double res=-INF;
   FO(i,2,n) {
       dp[i]=max(dp[i-1]+b[i],b[i]);
       res=max(res,dp[i]);
   }
   return sum<=res;
}
int main ()
{
    scanf("%d",&n);
    FOR(i,1,n) scanf("%d",a+i);
    double l=0, r=INF, mid;
    FOR(i,1,200) {
        mid=(l+r)/2.0;
        if (check(mid)) r=mid;
        else l=mid;
    }
    printf("%.3lf\n",mid);
    return 0;
}