hdu3613 Best Reward

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;
}