UVA 12594 Naming Babies

$dp$，斜率优化。

设$dp[j][i]$表示前$i$个位置分成$j$段的最小值，递推式很好写，预处理几个前缀和就可以了，然后斜率优化即可。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = ;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c))
    {
        x = x *  + c - '';
        c = getchar();
    }
}

int T,k;
char t[],s[];
int len,pos[];
int sum0[],sum1[],sum2[],dp[][];
int q[],f1,f2;

bool delete1(int t,int a,int b,int c)
{
    if( dp[t][b]-sum1[b]-b*sum0[c]+b*sum0[b]+sum0[b]+sum2[b] <=
        dp[t][a]-sum1[a]-a*sum0[c]+a*sum0[a]+sum0[a]+sum2[a]
    ) return ;
    return ;
}

bool delete2(int t,int a,int b,int c)
{
    if(
       ((dp[t][c]-sum1[c]+c*sum0[c]+sum0[c]+sum2[c])-(dp[t][b]-sum1[b]+b*sum0[b]+sum0[b]+sum2[b]))*(b-a)<=
       ((dp[t][b]-sum1[b]+b*sum0[b]+sum0[b]+sum2[b])-(dp[t][a]-sum1[a]+a*sum0[a]+sum0[a]+sum2[a]))*(c-b)
    ) return ;
    return ;
}

int main()
{
    scanf("%d",&T); int cas=;
    while(T--)
    {
        scanf("%s%d%s",t,&k,s);
        for(int i=;t[i];i++) pos[t[i]-'a']=i;

        sum0[]=sum1[]=sum2[]=;

        for(int i=;s[i];i++)
        {
            if(i>) sum0[i]=sum0[i-];
            sum0[i]=sum0[i]+pos[s[i]-'a'];

            if(i>) sum1[i]=sum1[i-];
            sum1[i]=sum1[i]+i*pos[s[i]-'a'];

            if(i>) sum2[i]=sum2[i-];
            sum2[i]=sum2[i]+pos[s[i]-'a']*pos[s[i]-'a'];
        }

        for(int i=;s[i];i++) dp[][i]=sum1[i]-sum2[i];

        len=strlen(s);

        for(int j=;j<=k;j++)
        {
            f1=f2=; q[]=j-;
            for(int i=j-;i<len;i++)
            {
                while()
                {
                    if(f2-f1+<) break;
                    if(delete1(j-,q[f1],q[f1+],i)) f1++;
                    else break;
                }

                dp[j][i]=dp[j-][q[f1]]+(sum1[i]-sum1[q[f1]])-(q[f1]+)*(sum0[i]-sum0[q[f1]])-(sum2[i]-sum2[q[f1]]);

                while()
                {
                    if(f2-f1+<) break;
                    if(delete2(j-,q[f2-],q[f2],i)) f2--;
                    else break;
                }

                f2++;
                q[f2]=i;
            }
        }

        printf("Case %d: %d\n",cas++,dp[k][len-]);
    }
    return ;
}