【KMP】 HDOJ 3613 Best Reward

做了很久，做不出来，只好看题解了。。。要注意到分割后的两个串是有特点的。前一个串一定包含原始串的第一个字符，后一个串一定包含原始串最后一个字符。如果能注意到就可以应该能想出来了。。。把原始串记为S1， 将原始串翻转记为S2。然后先将S1作为模式串，用S1匹配S2，最后得到的最大匹配就是S1串前缀最长回文串，利用next数组就可以求出所有S1串前缀回文串。。。同样的将S2作为模式串，用S2匹配S1，最后得到的最大匹配就是S1串后缀最长回文串，利用next数组就可以楸树所有S1串后缀回文串。。。。。。最后遍历一遍，算一算最大的价值就行了。。。

#include <iostream>#include <sstream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <bitset>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <climits>#define maxn 500005#define eps 1e-6#define mod 10007#define INF 99999999#define lowbit(x) (x&(-x))//#define lson o<<1, L, mid//#define rson o<<1 | 1, mid+1, Rtypedef long long LL;using namespace std;int v[30], len;char s1[maxn], s2[maxn];int next[maxn];bool pre[maxn];bool pos[maxn];int sum[maxn];void read(void){int i, j;for(i = 0; i < 26; i++) scanf("%d", &v[i]);scanf("%s", s1);len = strlen(s1);for(i = 0, j = len-1; i < len; i++, j--)s2[i] = s1[j];}void init(void){memset(pre, 0, sizeof pre);memset(pos, 0, sizeof pos);memset(sum, 0, sizeof sum);for(int i = 1; i <= len; i++)sum[i] = sum[i-1] + v[s1[i-1]-'a'];}void get(char *p){int i, j;next[0] = next[1] = 0;for(i = 1; i <= len; i++) {j = next[i];while(j && p[i] != p[j]) j = next[j];next[i+1] = p[i] == p[j] ? j + 1 : 0;}}int kmp(char *p, char *t){int i, j;for(i = 0, j = 0; i < len; i++) {while(j && t[i] != p[j]) j=next[j];if(t[i] == p[j]) j++;}return j;}void debug(void){for(int i = 1; i <= len; i++) {printf("%d\n", sum[i]);}}void work(void){int i, tmp, ans;get(s1);tmp = kmp(s1, s2);while(tmp) {pre[tmp] = true;tmp = next[tmp];}get(s2);tmp = kmp(s2, s1);while(tmp) {pos[len - tmp + 1] = true;tmp = next[tmp];}//debug();ans = -INF;for(i = 1; i < len; i++) {tmp = 0;if(pre[i]) tmp += sum[i];if(pos[i + 1]) tmp += sum[len] - sum[i];if(tmp > ans) ans=tmp;}printf("%d\n", ans);}int main(void){int _;while(scanf("%d", &_)!=EOF) {while(_--){read();init();work();}}return 0;}