【后缀数组】 HDOJ 5008 Boring String Problem
通过后缀数组可以找到第K的串，然后用二分找到最左边的编号。。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 40005
#define eps 1e-10
#define mod 1000000007
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
typedef long long LL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head
char s[maxn];
int sa[maxn], c[maxn], t1[maxn], t2[maxn];
int rank[maxn], height[maxn];
void build(int n, int m)
{
int *x = t1, *y = t2, p;
for(int i = 0; i < m; i++) c[i] = 0;
for(int i = 0; i < n; i++) c[x[i] = s[i]]++;
for(int i = 1; i < m; i++) c[i] += c[i-1];
for(int i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;
for(int k = 1; k <= n; k <<= 1) {
p = 0;
for(int i = n-k; i < n; i++) y[p++] = i;
for(int i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
for(int i = 0; i < m; i++) c[i] = 0;
for(int i = 0; i < n; i++) c[x[y[i]]]++;
for(int i = 1; i < m; i++) c[i] += c[i-1];
for(int i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
p = 1, swap(x, y), x[sa[0]] = 0;
for(int i = 1; i < n; i++)
x[sa[i]] = y[sa[i]] == y[sa[i-1]] && y[sa[i] + k] == y[sa[i-1] + k] ? p-1 : p++;
if(p >= n) break;
m = p;
}
}
void getheight(int n)
{
int k = 0;
for(int i = 0; i <= n; i++) rank[sa[i]] = i;
for(int i = 0; i < n; i++) {
if(k) k--;
int j = sa[rank[i] - 1];
while(s[i + k] == s[j + k]) k++;
height[rank[i]] = k;
}
}
int p[maxn];
LL st1[maxn][20], st2[maxn][20];
LL sum[maxn];
void init(int n, int *a, LL (*t)[20])
{
p[0] = -1;
for(int i = 1; i <= n; i++) p[i] = i & (i-1) ? p[i-1] : p[i-1] + 1;
for(int i = 1; i <= n; i++) t[i][0] = a[i];
for(int j = 1; j <= p[n]; j++)
for(int i = 1; i + (1 << j) - 1 <= n; i++)
t[i][j] = min(t[i][j-1], t[i + (1 << (j-1))][j-1]);
}
LL query(int l, int r, LL (*t)[20])
{
LL k = p[r - l + 1];
return min(t[l][k], t[r - (1 << k) + 1][k]);
}
LL lcp(int a, int b)
{
if(a > b) swap(a, b);
return query(a+1, b, st1);
}
int find(int n, LL x)
{
LL bot = 0, top = n, mid, res;
while(top >= bot) {
mid = (top + bot) >> 1;
if(sum[mid] >= x) top = mid - 1, res = mid;
else bot = mid + 1;
}
return res;
}
int search(int pos, int n, LL k)
{
int bot = pos + 1, top = n, mid, res = -1;
while(top >= bot) {
mid = (top + bot) >> 1;
if(lcp(pos, mid) < k) top = mid - 1;
else bot = mid + 1, res = mid;
}
if(res != -1) return query(pos, res, st2) + 1;
return sa[pos] + 1;
}
void work(void)
{
int n = strlen(s), l, r, kk, now, tmp;
LL k, v;
build(n+1, 128);
getheight(n);
init(n, height, st1);
init(n, sa, st2);
sum[0] = l = r = 0;
for(int i = 1; i <= n; i++) sum[i] = n - sa[i] - height[i];
for(int i = 1; i <= n; i++) sum[i] += sum[i-1];
scanf("%d", &kk);
while(kk--) {
scanf("%I64d", &v);
k = (l ^ r ^ v) + 1;
if(k > sum[n]) printf("0 0\n"), l = r = 0;
else {
now = find(n, k);
tmp = k - sum[now-1] + height[now];
l = min(sa[now] + 1, search(now, n, tmp));
r = l + tmp - 1;
printf("%d %d\n", l, r);
}
}
}
int main(void)
{
while(scanf("%s", s)!=EOF) work();
return 0;
}