【斜率优化】UVA 12594 Naming Babies

通道

题意：给出一个字符串和一个k，将字符串分成k段，每段的求法是sigma【(i − P OS) ∗ P OS】，问如何分段使得总和最小

思路：

设dp[i][j] 为将前j个字符分成i段的最小值。a[i]为第i个字符在原始字符串的位置。

那么dp[i][j] = min(dp[i][j],dp[i-1][k]+0*a[k+1]+1*a[k+2] + ...... + (j-k+1)*a[j] - a[k+1]^2 - ...... - a[j]^2)

==》k*a[k+1]+(k+1)*a[k+2] + ...... + (j-1)*a[j] - k*(a[k+1] + ....... + a[j]) - a[k+1]^2 - ...... - a[j]^2

用s1[]记录a[i]的和，s2[]记录(i-1)*a[i]的和，s3[]记录a[i]^2的和

然后就可以用斜率优化了

代码：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>

using namespace std;
const int maxm = 505;
const int maxn = 20005;

int dp[maxm][maxn];
int key[30],n,m,a[maxn],s1[maxn],s2[maxn],s3[maxn];
int q[maxn],hd,tl;
char str[maxn];

int getDP(int j,int k,int i){
    return dp[i-1][k] + s2[j] - s2[k] - k*(s1[j]-s1[k]) - (s3[j]-s3[k]);
}
int getY(int j,int k,int i){
    return (dp[i-1][k]-s2[k]+k*s1[k]+s3[k]) - (dp[i-1][j]-s2[j]+j*s1[j]+s3[j]);
}
int getX(int j,int k){
    return k - j;
}

int solve(){
    for(int j=1;j<=n;j++) dp[1][j] = s2[j] - s3[j];
    for(int i=2;i<=m;i++){
        hd = tl = 0;
        q[tl++] = i-1;
        for(int j=i;j<=n;j++){
            while(hd+1<tl && getY(q[hd],q[hd+1],i) <= s1[j]*getX(q[hd],q[hd+1])) hd++;
            dp[i][j] = getDP(j,q[hd],i);
            while(hd+1<tl && getY(q[tl-1],j,i)*getX(q[tl-2],q[tl-1])
                  <= getY(q[tl-2],q[tl-1],i)*getX(q[tl-1],j)) tl--;
            q[tl++] = j;
        }
    }
    return dp[m][n];
}

int main(){
    int cas;
    scanf("%d",&cas);
    for(int T = 1; T <= cas; T++){
        scanf("%s%d",str,&m);
        for(int i=0;i<strlen(str);i++){
            key[str[i]-'a'] = i;
        }
        scanf("%s",str);n = strlen(str);
        a[1] = key[str[0]-'a'];
        s1[1] = a[1];s2[1] = 0;s3[1] = a[1]*a[1];
        for(int i=1;i<n;i++){
            a[i+1] = key[str[i]-'a'];
            s1[i+1] = s1[i] + a[i+1];
            s2[i+1] = s2[i] + i*a[i+1];
            s3[i+1] = s3[i] + a[i+1]*a[i+1];
        }
        printf("Case %d: %d\n",T,solve());
    }

    return 0;
}