Description
Input
Output
一个数,最小方差乘以 m^2 后的值
Sample Input
1 2 5 8 6
Sample Output
HINT
1≤n≤3000,保证从 S 到 T 的总路程不超过 30000
/* 先吐槽一波数据,n=2146,只有2143个数的输入也是没谁了,害的我找了半天错误! 设m天走过的路程分别是a1,a2...am,平均数为p=dis[n]/m 化简一波式子 ans=Σ((ai-p)*(ai-p))*m*m=m*Σ(ai*ai)-dis[n]*dis[n] f[i][j]表示第i天走到j地的最小 Σai*ai 首先求出动态转移方程 f[i][j]=min(f[i][k]+(dis[j]-dis[k])^2) 可以斜率优化,化简上式可得,若k1比k2更优,则有: (dis[k1]^2-dis[k2]^2+f[i-1][k1]-f[i-1][k2])/(dis[k1]-dis[k2])>dis[j]*2 */ #include<cstdio> #include<iostream> #include<cstring> #define N 3010 #define lon long long using namespace std; lon dis[N],f[N][N];int n,m,q[N]; lon read(){ lon num=0,flag=1;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')flag=-1;c=getchar();} while(c>='0'&&c<='9'){num=num*10+c-'0';c=getchar();} return num*flag; } double lv(int i,int k1,int k2){ return double(dis[k1]*dis[k1]-dis[k2]*dis[k2]+f[i-1][k1]-f[i-1][k2])/double(dis[k1]-dis[k2]); } int main(){ scanf("%d%d",&n,&m); for(int i=1;i<=n;i++){ int x=0;scanf("%d",&x); dis[i]=dis[i-1]+(lon)x; } memset(f,127/3,sizeof(f)); int h,t;f[0][0]=0; for(int i=1;i<=m;i++){ h=0;t=0;q[0]=0; for(int j=1;j<=n;j++){ while(h<t&&lv(i,q[h+1],q[h])<=dis[j]*2) h++; f[i][j]=f[i-1][q[h]]+(dis[j]-dis[q[h]])*(dis[j]-dis[q[h]]); while(h<t&&lv(i,j,q[t])<=lv(i,q[t],q[t-1])) t--; q[++t]=j; } } cout<<f[m][n]*m-dis[n]*dis[n]; return 0; }