hdu3613 Best Reward

Posted 2020-10-16 poorpool

先manacher。然后前缀和价值，枚举切点，O(1)判断切后是否回文

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int T, val[1000005], p[1000005], len, w[255];
char a[1000005];
void manacher(){
    int mx=0, id=0;
    for(int i=1; i<len; i++){
        if(i<mx)    p[i] = min(p[2*id-i], mx-i);
        else    p[i] = 1;
        while(a[i-p[i]]==a[i+p[i]]) p[i]++;
        if(mx<i+p[i])   mx = i + p[i], id = i;
    }
}
int main(){
    cin>>T;
    w[‘#‘] = w[‘$‘] = 0;
    while(T--){
        for(int i=0; i<26; i++)
            scanf("%d", &w[i+‘a‘]);
        scanf("%s", a);
        len = strlen(a);
        for(int i=len; i>=0; i--){
            a[2*i+2] = a[i];
            a[2*i+1] = ‘#‘;
        }
        len = 2 * len + 1;
        a[0] = ‘$‘;
        manacher();
        for(int i=1; i<len; i++)
            val[i] = val[i-1] + w[(int)a[i]];
        int maxans=0;
        for(int i=3; i<len; i+=2){
            int tmp=0;
            if((i+1)/2+p[(i+1)/2]>i)    tmp += val[i];
            if((i+len)/2+p[(i+len)/2]>len)  tmp += val[len-1] - val[i-1];
            maxans = max(maxans, tmp);
        }
        printf("%d\n", maxans);
    }
    return 0;
}