HDU 5558 后缀数组+二分

题意有一些绕，但其实就是对于不断变化的i，求以j(0=j<i)使得suffix[j]与suffix[i]的最长公共前缀最长，如果有多个j，则取最小的j。

可以在rank数组中二分，在1-rank[i-1]中二分最接近i的j使得sa[j]小于i，通俗地说就是rank比的rank[i]小，并且位于i之前的后缀。因为这个是左边最接近rank[i]的，所以与suffix[i]的最长公共前缀一定是满足最大的。接下来再根据得到的LCP值，二分一个最小的j。同理，再从rank[i+1]到rank[n]中作类似二分。两者结合得到当前的K和T。

其实这道题中间过程也可以不用二分，直接向左右暴力扫。速度会快很多。

#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=1e5+10;


int MIN(int a,int b) {
    return a<b?a:b;
}
int MAX(int a,int b) {
    return a>b?a:b;
}

int val[N];
struct RMQ {
    int dp[N][22];
    int (*cmp) (int,int);
    void setMin() {
        cmp=MIN;
    }
    void setMax() {
        cmp=MAX;
    }
    void init(int n,int *val) {
        for (int i=0; i<=n; i++)
            dp[i][0]=val[i];
        for (int j=1; (1<<j)<=n; j++) {
            int k=1<<(j-1);
            for (int i=0; i+k<=n; i++)
                dp[i][j]=cmp(dp[i][j-1],dp[i+k][j-1]);
        }
    }
    int query(int a,int b) {
        if (a>b) swap(a,b);
        int dis=b-a+1;
        int k=log((double)dis)/log(2.0);
        return cmp(dp[a][k],dp[b-(1<<k)+1][k]);
    }
} rmq;

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 = 0; i < m; ++i) cnt[i] = 0;
        for (i = 0; i < n; ++i) cnt[x[i]=r[i]]++;
        for (i = 1; i < m; ++i) cnt[i] += cnt[i-1];
        for (i = n-1; i >= 0; --i) sa[--cnt[x[i]]] = i;
        for (j = 1, p = 1; p < n; j *= 2, m = p) {
            for (p = 0, i = n - j; i < n; ++i) y[p++] = i;
            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) cnt[i] = 0;
            for (i = 0; i < n; ++i) cnt[wv[i]]++;
            for (i = 1; i < m; ++i) cnt[i] += cnt[i-1];
            for (i = n-1; i >= 0; --i) sa[--cnt[wv[i]]] = y[i];
            for (swap(x, y), 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++;
        }
    }
    void calcHeight(char r[], int n) {
        int i, j, k = 0;
        for (i = 1; i <= n; ++i) rk[sa[i]] = i;
        for (i = 0; i < n; height[rk[i++]] = k)
            for (k?k--:0, j = sa[rk[i]-1]; 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+1,rb);
    }
    int st[N][22];
    int preLog2[N];
    void initST(int n) {
        for (int i=1; i<=n; i++)
            st[i][0]=height[i];
        for (int j=1; (1<<j)<=n; j++) {
            int k=1<<(j-1);
            for (int i=1; i+k<=n; i++)
                st[i][j]=min(st[i][j-1],st[i+k][j-1]);
        }
        preLog2[1]=0;
        for(int i=2;i<=n;i++){
            preLog2[i]=preLog2[i-1]+((i&(i-1))==0);
        }
    }
    int queryST(int a,int b) {
        if (a>b) swap(a,b);
        int dis=b-a+1;
        int k=preLog2[dis];
        return min(st[a][k],st[b-(1<<k)+1][k]);
    }
    void solve(int n) {
        calcSA(s,n+1,128);
        calcHeight(s,n);
        initST(n);
        rmq.setMin();
        rmq.init(n,sa);
        printf("-1 %d\n",s[0]);
        int i=1;
        while (i<n) {
            int l=1,r=rk[i]-1;
            int k=-1,t=s[i];
            int minIndex=-1;
            while (l<=r) {
                int mid=(l+r)>>1;
                int minSA=rmq.query(mid,rk[i]-1);
                if (minSA<i)
                    l=mid+1,minIndex=minSA;
                else
                    r=mid-1;
            }
            if (minIndex!=-1) {
                int u=lcp(minIndex,i,n);
                if (u) {
                    k=u;
                    int l=1,r=rk[i]-1,ret=-1;
                    while (l<=r) {
                        int mid=(l+r)>>1;
                        int curLCP=lcp(sa[mid],i,n);
                        if (curLCP>=u)
                            r=mid-1,ret=mid;
                        else l=mid+1;
                    }
                    t=rmq.query(ret,rk[i]-1);
                }
            }
            l=rk[i]+1;
            r=n;
            minIndex=-1;
            while (l<=r) {
                int mid=(l+r)>>1;
                int minSA=rmq.query(rk[i]+1,mid);
                if (minSA<i)
                    r=mid-1,minIndex=minSA;
                else
                    l=mid+1;
            }
            if (minIndex!=-1) {
                int u=lcp(minIndex,i,n);
                if (u&&u>=k) {
                    if (u==k)
                        t=min(t,minIndex);
                    else
                        t=minIndex;
                    k=u;
                    int l=rk[i]+1,r=n,ret=-1;
                    while (l<=r) {
                        int mid=(l+r)>>1;
                        int curLCP=lcp(sa[mid],i,n);
                        if (curLCP>=u)
                            l=mid+1,ret=mid;
                        else r=mid-1;
                    }
                    t=min(t,rmq.query(rk[i]+1,ret));
                }
            }
            if (k==-1) {
                printf("-1 %d\n",s[i]);
                i++;
            } else {
                printf("%d %d\n",k,t);
                i+=k;
            }
        }
    }
} suf;

int main () {
    int T;
    scanf("%d",&T);
    while (T--) {
        scanf("%s",s);
        int n=strlen(s);
        static int cas=1;
        printf("Case #%d:\n",cas++);
        suf.solve(n);
    }
    return 0;
}