HDU 3948 不同回文子串个数

集训队论文中有求不同子串个数的做法，就是扫一遍height数组，过程中根据height数组进行去重。对于本题也是雷同的，只是每一次不是根据与排名在上一位的LCP去重，而是与上一次统计对答案有贡献的后缀进行比较去重。

几组数据

abacaba 7

abbacaa 7

baabcaa 5

 #include <iostream>
 #include <vector>
 #include <algorithm>
 #include <string>
 #include <string.h>
 #include <stdio.h>
 #include <queue>
 #include <stack>
 #include <map>
 #include <set>
 #include <cmath>
 #include <ctime>
 #include <cassert>
 #include <sstream>
 using namespace std;

 const int N=*;

 char s[N];
 struct SuffixArray {
      int wa[N], wb[N], cnt[N], wv[N];
     int rk[N], height[N];
     int sa[N];
     bool cmp(int r[], int a, int b, int l) {
         return r[a] == r[b] && r[a+l] == r[b+l];
     }
     void calcSA(char r[], int n, int m) {
         int i, j, p, *x = wa, *y = wb;
         for (i = ; i < m; ++i) cnt[i] = ;
         for (i = ; i < n; ++i) cnt[x[i]=r[i]]++;
         for (i = ; i < m; ++i) cnt[i] += cnt[i-];
         for (i = n-; i >= ; --i) sa[--cnt[x[i]]] = i;
         for (j = , p = ; p < n; j *= , m = p) {
             for (p = , i = n - j; i < n; ++i) y[p++] = i;
             for (i = ; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
             for (i = ; i < n; ++i) wv[i] = x[y[i]];
             for (i = ; i < m; ++i) cnt[i] = ;
             for (i = ; i < n; ++i) cnt[wv[i]]++;
             for (i = ; i < m; ++i) cnt[i] += cnt[i-];
             for (i = n-; i >= ; --i) sa[--cnt[wv[i]]] = y[i];
             for (swap(x, y), p = , x[sa[]] = , i = ; i < n; ++i)
                 x[sa[i]] = cmp(y, sa[i-], sa[i], j) ? p- : p++;
         }
     }
     void calcHeight(char r[], int n) {
         int i, j, k = ;
         for (i = ; i <= n; ++i) rk[sa[i]] = i;
         for (i = ; i < n; height[rk[i++]] = k)
             for (k?k--:, j = sa[rk[i]-]; r[i+k] == r[j+k]; k++);
     }
     int lcp(int a,int b,int len) {
         if (a==b) return len-a;
         int ra=rk[a],rb=rk[b];
         if (ra>rb) swap(ra,rb);
         return queryST(ra+,rb);
     }
     int st[N][];
     void initST(int n) {
         for (int i=; i<=n; i++)
             st[i][]=height[i];
         for (int j=; (<<j)<=n; j++) {
             int k=<<(j-);
             for (int i=; i+k<=n; i++)
                 st[i][j]=min(st[i][j-],st[i+k][j-]);
         }
     }
     int queryST(int a,int b) {
         if (a>b) swap(a,b);
         int dis=b-a+;
         int k=log((double)dis)/log(2.0);
         return min(st[a][k],st[b-(<<k)+][k]);
     }
     void solve(int cas) {
         int n=strlen(s);
         s[n]='#';
         for (int i=;i<n;i++) {
             s[n++i]=s[n--i];
         }
         int o=n;
         n=*n+;
         s[n]='\0';
         calcSA(s,n+,);
         calcHeight(s,n);
         initST(n);
         long long ret=;
         int curLcp=;
         for (int i=;i<=n;i++) { //odd
             int pos=sa[i];
             curLcp=min(curLcp,height[i]);
             if (pos<o) {
                 int ops=n--pos;
                 int now=lcp(pos,ops,n);
                 ret+=max(,now-curLcp);
                 if (now>=curLcp)
                     curLcp=now;
             }
         }
         curLcp=;
         for (int i=;i<=n;i++) { //even
             int pos=sa[i];
             curLcp=min(curLcp,height[i]);
             if (pos<o) {
                 int ops=n-pos;
                 int now=lcp(pos,ops,n);
                 ret+=max(,now-curLcp);
                 if (now>=curLcp)
                     curLcp=now;
             }
         }
         printf("Case #%d: %I64d\n",cas,ret);
     }
 }suf;

 int main () {
     //freopen("out.txt","r",stdin);
     int T;
     scanf("%d",&T);
     int cas=;
     while (T--) {
        scanf("%s",s);
        suf.solve(cas);
        cas++;
     }
     return ;
 }