UVA11584 Partitioning by Palindromes

题意：一个字符串(l<1000),问最少能分多少个回文串

题解：dp[i]代表前i个字符串最少的回文串，可以由前面的dp递推，dp[i] = max(dp[i],dp[i-t]+t);(1<t<=i)这里是n^2，判断回文串用Manacher

#include <bits/stdc++.h>
#define maxn 100100
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
char s[maxn], s1[maxn];
int p[2*maxn], dp[maxn];
int init(){
    int j = 2, l=strlen(s);
    s1[0] = '&';s1[1] = '#';
    for(int i=0;i<l;i++){
        s1[j++] = s[i];
        s1[j++] = '#';
    }
    s1[j] = 0;
    return j;
}
void manacher(char *s){
    int id=0,ma=0,mlen=-1;
    int len = init();
    for(int i=0;i<len;i++){
        if(i<ma) p[i] = min(p[2*id-i], ma-i);
        else p[i] = 1;
        while(s1[i-p[i]] == s1[i+p[i]]) p[i]++;
        if(i+p[i] > ma){
            id = i;
            ma = i+p[i];
        }
    }
}
int main(){
    int T, l;
    scanf("%d", &T);
    while(T--){
        scanf("%s", s);
        l = strlen(s);
        manacher(s);
        for(int i=2;i<=2*l;i+=2){
            dp[i] = i/2;
            for(int j=0;j<i;j+=2){
                if(p[(i+j+2)/2]-1 >= (i-j)/2) dp[i] = min(dp[i], dp[j]+1);
            }
        }
        printf("%d\n", dp[l*2]);
    }
    return 0;
}