HDU3613 Best Reward（扩展KMP）

题意：给一个由小写字母构成的字符串，每个字母否有对应的值，将串分成两个子串，如果子串是回文串的话子串的值就是每个字符的值之和；否则子串的值就是0.原串的值等于两个子串之和，求原串能取到的最大的值。

思路：求回文子串的和的最大值显然需要找出所有可能的回文子串，可以用扩展KMP。先将原串s1反转得到s2，以s1为目标串s2为模式串做一次扩展KMP得到extend1数组，此时若 extend1[i] == len - i ,则后面len - i 个字符一定构成了回文子串，此时需要判断前 i 个字符是否构成了回文串，只需将s2作为目标串，s1作为模式串，再做一次扩展KMP，得到数组extend2.若 extend2[len - i] == i ，则前i个字符也一定构成了回文串。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<vector>#include<map>#include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;const int mod = 1000000007;const int maxn=500005;typedef long long ll;int nxt[maxn];//获得子串的next数组void get_next(char str[]){    int i = 0, j, pos;    int len = strlen(str);    nxt[0] = len;    while(i + 1 < len && str[i] == str[i + 1]){        ++i;    }    nxt[1] = i;    pos = 1;    for(i=2; i<len; ++i){        if(nxt[i - pos] < pos + nxt[pos] - i){            nxt[i] = nxt[i - pos];        } else {            j = nxt[pos] + pos - i;            if(j < 0){                j = 0;            }            while(i + j < len && str[i + j] == str[j]){                ++j;            }            nxt[i] = j;            pos = i;        }    }}//计算extend数组void EXKMP(char target[], char pattern[], int extend[]){    int i = 0, j, pos;    get_next(pattern);    int len1 = strlen(target);    int len2 = strlen(pattern);    while(i < len1 && i < len2 && target[i] == pattern[i]){        ++i;    }    extend[0] = i;    pos = 0;    for(i=1; i<len1; ++i){        if(nxt[i - pos] < pos + extend[pos] - i){            extend[i] = nxt[i - pos];        } else {            j = extend[pos] + pos - i;            if(j < 0){                j = 0;//从头匹配            }            while(i + j < len1 && j < len2 && target[i + j] == pattern[j]){                ++j;            }            extend[i] = j;            pos = i;//更新pos        }    }}int sum[maxn], val[30];char s1[maxn], s2[maxn];int extend1[maxn], extend2[maxn];int main(){    int t;    scanf("%d", &t);    while(t--){        for(int i=0; i<26; ++i){            scanf("%d", &val[i]);        }        scanf("%s", s1);        int len = strlen(s1);        for(int i=0; i<len; ++i){            s2[i] = s1[len - i - 1];        }        s2[len] = 0;        sum[0] = val[s1[0] - 'a'];        for(int i=1; i<len; ++i){            sum[i] = sum[i - 1] + val[s1[i] - 'a'];        }        EXKMP(s1, s2, extend1);        EXKMP(s2, s1, extend2);        int ans = 0, now = 0;        for(int i=1; i<len; ++i){//从1开始扫避免整个串是回文串的情况            if(extend1[i] == len - i){//后面len-i个字符构成了回文串                now += sum[len - 1] - sum[i - 1];            }            if(extend2[len - i] == i){//前面i个字符构成了回文串                now += sum[i - 1];            }            if(now > ans){                ans = now;            }            now = 0;        }        printf("%d\n", ans);    }    return 0;}/*31 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1aaaabc1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1aba1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1acacac*/