hdu 5008 Boring String Problem

      后缀数组+RMQ+二分。 先跑后缀数组，然后求出一组非优解。然后利用height数组，求出能搞出满足串的rank范围(二分RMQ，求出右端点)，然后在rank范围RMQ一下sa数组，找出最小的左端点就可以了。

      后缀数组模板摘自：http://blog.csdn.net/dr5459/article/details/9051149

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<queue>
#include<cmath>
#include<stack>
#include<set>
#include<map>
#define inf 0x3f3f3f3f
#define CLR(a, b) memset(a, b, sizeof(a))
#define lowbit(x) (x&(-x))
using namespace std;

#ifdef __int64
typedef __int64 LL;
#else
typedef long long LL;
#endif

const int maxn=211111;
/******************************************************************
** 后缀数组 Suffix Array
** INIT：solver.call_fun(char* s);
** SP_USE: solver.LCS(char *s1,char* s2); //最长公共字串
******************************************************************/
struct SuffixArray
{
int r[maxn];
int sa[maxn],rank[maxn],height[maxn];
int t[maxn],t2[maxn],c[maxn],n;
int m;///模板长度
void init(char* s)
{
n=strlen(s);
for (int i=0; i<n; i++) r[i]=int(s[i]);
m=300;
}
int cmp(int *r,int a,int b,int l)
{
return r[a]==r[b]&&r[a+l]==r[b+l];
}
/**
字符要先转化为正整数
待排序的字符串放在r[]数组中，从r[0]到r[n-1]，长度为n，且最大值小于m。
所有的r[i]都大于0,r
无意义算法中置0
函数结束后，结果放在sa[]数组中(名次从1..n)，从sa[1]到sa
。s[0]无意义
**/
void build_sa()
{
int i,k,p,*x=t,*y=t2;
r[n++]=0;
for (i=0; i<m; i++) c[i]=0;
for (i=0; i<n; i++) c[x[i]=r[i]]++;
for (i=1; i<m; i++) c[i]+=c[i-1];
for (i=n-1; i>=0; i--) sa[--c[x[i]]]=i;
for (k=1,p=1; k<n; k*=2,m=p)
{
for (p=0,i=n-k; i<n; i++) y[p++]=i;
for (i=0; i<n; i++) if (sa[i]>=k) y[p++]=sa[i]-k;
for (i=0; i<m; i++) c[i]=0;
for (i=0; i<n; i++) c[x[y[i]]]++;
for (i=1; i<m; i++) c[i]+=c[i-1];
for (i=n-1; i>=0; i--) sa[--c[x[y[i]]]]=y[i];
swap(x,y);
p=1;
x[sa[0]]=0;
for (i=1; i<n; i++) x[sa[i]]=cmp(y,sa[i-1],sa[i],k)?p-1:p++;
}
n--;
}
/**
height[2..n]:height[i]保存的是lcp(sa[i],sa[i-1])
rank[0..n-1]:rank[i]保存的是原串中suffix[i]的名次
**/
void getHeight()
{
int i,j,k=0;
for (i=1; i<=n; i++) rank[sa[i]]=i;
for (i=0; i<n; i++)
{
if (k) k--;
j=sa[rank[i]-1];
while (r[i+k]==r[j+k]) k++;
height[rank[i]]=k;
}
}
int d[maxn][20];
///元素从1编号到n
void RMQ_init(int A[],int n)
{
for (int i=1; i<=n; i++) d[i][0]=A[i];
for (int j=1; (1<<j)<=n; j++)
for (int i=1; i+(1<<(j-1))<=n; i++)
d[i][j]=min(d[i][j-1],d[i+(1<<(j-1))][j-1]);
}
int RMQ(int L,int R)
{
int k=0;
if(L>R) swap(L,R);
if(L == R) return n - sa[L];
L++;
// k = log2(R - L + 1);
while ((1<<(k+1))<=R-L+1) k++;
return min(d[L][k],d[R-(1<<k)+1][k]);
}

int d2[maxn][20];
void RMQ_init2(int A[],int n)
{
for (int i=1; i<=n; i++) d2[i][0]=A[i];
for (int j=1; (1<<j)<=n; j++)
for (int i=1; i+(1<<(j-1))<=n; i++)
d2[i][j]=min(d2[i][j-1],d2[i+(1<<(j-1))][j-1]);
}
int RMQ2(int L,int R)
{
int k=0;
if(L>R) swap(L,R);
// k = log2(R - L + 1);
while ((1<<(k+1))<=R-L+1) k++;
return min(d2[L][k],d2[R-(1<<k)+1][k]);
}
void LCP_init()
{
RMQ_init(height,n);
RMQ_init2(sa,n);
}
void call_fun(char* s)
{
init(s);//初始化后缀数组
build_sa();//构造后缀数组sa
getHeight();//计算height与rank
LCP_init();//初始化RMQ
}
LL ans[maxn];
int solve(int n)
{
LL tot=0;
for(int i=1; i<=n; i++)
{
tot+=(n-sa[i]-height[i]);
ans[i] = tot;
}
return tot;
}

void query(LL k, int &L, int &R)
{
int l = 1, r = n;
if(ans
< k)
{
L = 0; R = 0;
return ;
}
while(l <= r)
{
int m = (l + r) >> 1;
if(ans[m] <= k) l = m + 1;
else r = m - 1;
}
if(k == ans[r])
{
L = sa[r] + 1; R = n;
}
else
{
l = sa[r + 1] + 1;
k -= ans[r];
k --;
r = l + k + height[r + 1];
L = l; R = r;
}
l = rank[L - 1]; r = n;
while(l <= r)
{
int m = (l + r) >> 1;
if(RMQ(rank[L - 1], m) >= R - L + 1)l = m + 1;
else r = m - 1;
}
l = RMQ2(rank[L - 1], r);
R = R - L + l + 1;
L = l + 1;
}
} sol;

char ch[maxn];

int main()
{
while(scanf("%s",ch) != EOF)
{
int n=strlen(ch);
sol.call_fun(ch);
sol.solve(n);
int q;
scanf("%d", &q);
int l = 0, r = 0;
while(q --)
{
LL k;
scanf("%I64d", &k);
LL ll = l, rr = r, kk = k;
k = ll ^ rr ^ k;
sol.query(k + 1, l, r);
printf("%d %d\n", l, r);
}
}
return 0;
}