hdu 3948 The Number of Palindromes

后缀数组，求不同回文子串的数量

#include <STDIO.H>#include <STRING.H>// hdu 3948 const int MAXN = 200010;int wa[MAXN], wb[MAXN], ws[MAXN], wv[MAXN]; int rank[MAXN], height[MAXN], r[MAXN], sa[MAXN];inline int min(const int a, const int b){return a<b?a:b;}inline void swap(int &x, int &y) {int a = x; x = y; y = a;}bool cmp(int *r, int a, int b, int l){return (r[a] == r[b] && r[a+l] == r[b+l]);}// r： 待排序的字符串，长度为n，最大值小于m // ws：基数排序// sa：存放排序结果 void da(int *r, int *sa, int n, int m){//用倍增算法求后缀数组int i, j, p, *x = wa, *y = wb, *t;// 使用基数排序对长度为1的字符串进行排序for (i = 0; i< m; ++i) ws[i] = 0;// 初始化wsfor (i = 0; i< n; i++) ws[x[i]=r[i]]++;  for (i = 1; i< m; i++) ws[i]+=ws[i-1];for (i = n-1; i>= 0; --i) sa[--ws[x[i]]] = i; // 进行若干次基数排序，基数排序要分两次，第一次是对第二关键字排序// 第二次是对第一关键字排序。对第二关键字排序的结果可以利用上一次求得的sa直接算出for (j=1, p=1; p< n; j*=2, m=p){for (p=0, i=n-j; i< n; ++i) y[p++] = i;//变量j是当前字符串的长度，数组y保存的是对第二关键字排序的结果for (i=0; i< n; ++i) if (sa[i] >= j) y[p++] = sa[i] - j; for (i=0; i< n; ++i) wv[i] = x[y[i]];for (i=0; i< m; ++i) ws[i] = 0;for (i=0; i< n; ++i) ws[wv[i]]++;for (i=1; i< m; ++i) ws[i]+= ws[i-1];for (i=n-1; i>=0; --i) sa[--ws[wv[i]]] = y[i];// 计算rank 值for (t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i< n; ++i)x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;}}/*height：定义height[i]=suffix(sa[i-1])和suffix(sa[i])的最长公共前缀，也就是排名相邻的两个后缀的最长公共前缀*/void calHeight(int *r, int *sa, int n){int i, j, k = 0;for (i = 1; i<= n; ++i) rank[sa[i]] = i;for (i = 0; i< n; height[rank[i++]] = k)for (k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);}char str[MAXN];// 将字符串翻转后用一个特殊字符间隔连接到原字符串后int mGet() // r = str + '*' + reverse(str){int l;for (l = 0; str[l]; l++)r[l] = str[l];r[l] = '#';for (int i = l-1; i >= 0; i--)r[++l] = str[i];r[++l] = 0;return l;}int LOG[MAXN];int st[MAXN][20];/*st[i][j]表示height[j]到height[j+2^i-1]的最小值初始化st[i][0]=height[i]，st[i][j]=min(st[i][j-1],st[i+(1<<j-1)][j-1])*/void initRMQ(const int &n){//初始化RMQint i, j, k, limit ;for (i=0; i< n; ++i) st[i][0] = height[i];k = LOG[n];for (j = 1; j<= k; ++j){limit = n - (1<<(j-1));for (i = 0; i<= limit; ++i)st[i][j] = min(st[i][j-1], st[i+(1<<(j-1))][j-1]);}}int query(int x, int y){if (x > y) swap(x, y);++x;int k= LOG[y-x+1];return min(st[x][k], st[y-(1<<k)+1][k]);}int solve(int n){int ans = 0, st1 = 0, st2 = 0;int j;rank[n] = 0;for (int i = 2; i<= n; ++i){st1 = min(st1, height[i]);j = query(i, rank[n-sa[i]-1]);if (j > st1)ans += j-st1, st1 = j;st2 = min(st2, height[i]);j = query(i, rank[n-sa[i]]);if ( j > st2)ans += j-st2, st2 = j;}return ans;}int main()   {#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endifint t, cs = 0;int i;// LOG[i] == log2 i ；for (i = 1, LOG[0] = -1; i< MAXN; ++i)LOG[i] = LOG[i>>1] + 1;scanf("%d", &t);while (t--){scanf("%s", str);int n = mGet();da(r, sa, n+1, 128);calHeight(r, sa, n);initRMQ(n+1);printf("Case #%d: %d\n", ++cs, solve(n));}return 0;}