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

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  1 #include <iostream>
  2 #include <vector>
  3 #include <algorithm>
  4 #include <string>
  5 #include <string.h>
  6 #include <stdio.h>
  7 #include <queue>
  8 #include <stack>
  9 #include <map>
 10 #include <set>
 11 #include <cmath>
 12 #include <ctime>
 13 #include <cassert>
 14 #include <sstream>
 15 using namespace std;
 16 
 17 const int N=123456*2;
 18 
 19 char s[N];
 20 struct SuffixArray {
 21      int wa[N], wb[N], cnt[N], wv[N];
 22     int rk[N], height[N];
 23     int sa[N];
 24     bool cmp(int r[], int a, int b, int l) {
 25         return r[a] == r[b] && r[a+l] == r[b+l];
 26     }
 27     void calcSA(char r[], int n, int m) {
 28         int i, j, p, *x = wa, *y = wb;
 29         for (i = 0; i < m; ++i) cnt[i] = 0;
 30         for (i = 0; i < n; ++i) cnt[x[i]=r[i]]++;
 31         for (i = 1; i < m; ++i) cnt[i] += cnt[i-1];
 32         for (i = n-1; i >= 0; --i) sa[--cnt[x[i]]] = i;
 33         for (j = 1, p = 1; p < n; j *= 2, m = p) {
 34             for (p = 0, i = n - j; i < n; ++i) y[p++] = i;
 35             for (i = 0; i < n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j;
 36             for (i = 0; i < n; ++i) wv[i] = x[y[i]];
 37             for (i = 0; i < m; ++i) cnt[i] = 0;
 38             for (i = 0; i < n; ++i) cnt[wv[i]]++;
 39             for (i = 1; i < m; ++i) cnt[i] += cnt[i-1];
 40             for (i = n-1; i >= 0; --i) sa[--cnt[wv[i]]] = y[i];
 41             for (swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; ++i)
 42                 x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
 43         }
 44     }
 45     void calcHeight(char r[], int n) {
 46         int i, j, k = 0;
 47         for (i = 0; i <= n; ++i) rk[sa[i]] = i;
 48         for (i = 0; i < n; height[rk[i++]] = k)
 49             for (k?k--:0, j = sa[rk[i]-1]; r[i+k] == r[j+k]; k++);
 50     }
 51     int lcp(int a,int b,int len) {
 52         if (a==b) return len-a;
 53         int ra=rk[a],rb=rk[b];
 54         if (ra>rb) swap(ra,rb);
 55         return queryST(ra+1,rb);
 56     }
 57     int st[N][22];
 58     void initST(int n) {
 59         for (int i=1; i<=n; i++)
 60             st[i][0]=height[i];
 61         for (int j=1; (1<<j)<=n; j++) {
 62             int k=1<<(j-1);
 63             for (int i=1; i+k<=n; i++)
 64                 st[i][j]=min(st[i][j-1],st[i+k][j-1]);
 65         }
 66     }
 67     int queryST(int a,int b) {
 68         if (a>b) swap(a,b);
 69         int dis=b-a+1;
 70         int k=log((double)dis)/log(2.0);
 71         return min(st[a][k],st[b-(1<<k)+1][k]);
 72     }
 73     void solve(int cas) {
 74         int n=strlen(s);
 75         s[n]='#';
 76         for (int i=0;i<n;i++) {
 77             s[n+1+i]=s[n-1-i];
 78         }
 79         int o=n;
 80         n=2*n+1;
 81         s[n]='\0';
 82         calcSA(s,n+1,128);
 83         calcHeight(s,n);
 84         initST(n);
 85         long long ret=0;
 86         int curLcp=0;
 87         for (int i=1;i<=n;i++) { //odd
 88             int pos=sa[i];
 89             curLcp=min(curLcp,height[i]);
 90             if (pos<o) {
 91                 int ops=n-1-pos;
 92                 int now=lcp(pos,ops,n);
 93                 ret+=max(0,now-curLcp);
 94                 if (now>=curLcp)
 95                     curLcp=now;
 96             }
 97         }
 98         curLcp=0;
 99         for (int i=1;i<=n;i++) { //even
100             int pos=sa[i];
101             curLcp=min(curLcp,height[i]);
102             if (pos<o) {
103                 int ops=n-pos;
104                 int now=lcp(pos,ops,n);
105                 ret+=max(0,now-curLcp);
106                 if (now>=curLcp)
107                     curLcp=now;
108             }
109         }
110         printf("Case #%d: %I64d\n",cas,ret);
111     }
112 }suf;
113 
114 int main () {
115     //freopen("out.txt","r",stdin);
116     int T;
117     scanf("%d",&T);
118     int cas=1;
119     while (T--) {
120        scanf("%s",s);
121        suf.solve(cas);
122        cas++;
123     }
124     return 0;
125 }

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