HDU 2829 Lawrence （斜率DP）

斜率DP

设dp[i][j]表示前i点，炸掉j条边的最小值。j<i

dp[i][j]=min{dp[k][j-1]+cost[k+1][i]}

又由得出cost[1][i]=cost[1][k]+cost[k+1][i]+sum[k]*(sum[i]-sum[k])

cost[k+1][i]=cost[1][i]-cost[1][k]-sum[k]*(sum[i]-sum[k])

代入DP方程

可以得出 y=dp[k][j-1]-cost[1][k]+sum[k]^2

x=sum[k].

斜率sum[i]

const maxn=2008;
var n,m,i,j,h,t:longint;
    a,sum,cost,q:array[0..maxn*10] of int64;
    f:array[0..maxn,0..maxn] of int64;
function kx(x,y:int64):int64;
begin
    exit(sum[x]-sum[y]);
end;
function ky(x,y:int64):int64;
begin
    exit((f[x,j-1]-cost[x]+sum[x]*sum[x])-(f[y,j-1]-cost[y]+sum[y]*sum[y]))
end;
begin
    readln(n,m);
    for i:=1 to n do read(a[i]);
    for i:=1 to n do sum[i]:=sum[i-1]+a[i];
    for i:=1 to n do cost[i]:=cost[i-1]+a[i]*sum[i-1];
    for i:=1 to n do f[i,0]:=cost[i];
    for i:=1 to n do f[i,i-1]:=0;
    for j:=1 to m do
    begin
        h:=0; t:=1; q[1]:=j;
        for i:=j+1 to n do
        begin
            while (h<t) and (kx(q[h+1],q[h])*sum[i]>ky(q[h+1],q[h])) do inc(h);
            f[i,j]:=-sum[i]*sum[q[h]]+f[q[h],j-1]-cost[q[h]]+sum[q[h]]*sum[q[h]]+cost[i];
            while (h<t) and (ky(i,q[t])*kx(q[t],q[t-1])<=ky(q[t],q[t-1])*kx(i,q[t])) do dec(t);
            inc(t);
            q[t]:=i;
       end;
    end;
    writeln(f[n,m]);
end.