HDU 4691 front compression (suffix array LCA)

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4691

The question can only be TLE or MLE.

If the incorrect question is the prefix of all the previous inserts, use the suffix array TLE

If trie MLE is used, the problem cannot be solved even if it is wrong.

In fact, the positive solution is similar to the above input.

Two Methods: suffix array combined with rmq brute force water

When using rmq, note that N is included during initialization.

L ++ during query, because height is the current and the longest prefix.

Suffix array:

# Include <iostream> # include <stdio. h> # include <algorithm> # include <math. h> # include <string. h> using namespace STD; # define maxn 110000 # define ws ws1int wa [maxn], WB [maxn], WV [maxn], WS [maxn]; int CMP (int * r, int A, int B, int L) {return R [a] = R [B] & R [A + L] = R [B + L];} void da (char * r, int * Sa, int N, int m) {int I, j, P, * x = wa, * Y = WB, * t; for (I = 0; I <m; I ++) WS [I] = 0; for (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; 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 ++) 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]; 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 ++;} return;} int rank [maxn], height [maxn]; void calheight (char * r, int * Sa, int N) {int I, j, k = 0; for (I = 1; I <= N; I ++) rank [SA [I] = I; // calculate the rank when calculating the height, and use 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 ++); return ;} char rec [maxn]; int SA [maxn]; int N; int rmq [maxn] [20]; int rmq (int n) {int I, J; for (I = 0; I <= N; I ++) rmq [I] [0] = I; for (j = 1; (1 <j) <= N; j ++) for (I = 0; I + (1 <j)-1 <= N; I ++) {If (height [rmq [I] [J-1] <= height [rmq [I + (1 <(J-1)] [J-1]) rmq [I] [J] = rmq [I] [J-1]; elsermq [I] [J] = rmq [I + (1 <(J-1)] [J-1];} return 0;} int query (int l, int R) {If (L> r) L ^ = R, R ^ = L, L ^ = R; if (L! = R) l ++; int K = Log (double (R-l + 1)/log (2.0 ); if (height [rmq [l] [k] <= height [rmq [R-(1 <k) + 1] [k]) return rmq [l] [k]; elsereturn rmq [R-(1 <k) + 1] [k];} int find_ans (int K) {If (k <10) return 1; int ans = 0; while (k) {ans ++; k/= 10;} return ans;} int main () {int I, J, K, a, B, c, d, num; int mid, M; long ans1, ans2; while (scanf ("% s ", REC )! = EOF) {n = strlen (REC); ans1 = ans2 = 0; DA (REC, SA, N + 1,200 ); // here is n + 1 because the function contains <n calheight (REC, SA, N ); // note that N is used here. Because N is used in the function, note the following: // for (I = 0; I <n; I ++) // printf ("% d", height [I]); // printf ("\ n"); rmq (n); scanf ("% d ", & M); scanf ("% d", & A, & B); C = A, D = B; ans1 = B-A + 1; ans2 = B-A + 2 + 1; for (I = 1; I <m; I ++) {scanf ("% d", &, & B); If (rank [a]! = Rank [c]) {num = query (rank [a], rank [c]); num = height [num]; If (Num> B-) num = B-A; If (Num> D-C) num = D-C;} else {num = min (B-a, d-C );} ans1 + = B-A + 1; ans2 + = (B-a-num + 2) + find_ans (Num); C = A, D = B; // printf ("Ans: % d \ n", num);} printf ("% i64d % i64d \ n", ans1, ans2);} return 0 ;}

Brute force:

#include <iostream>#include <string.h>#include <stdio.h>#include <cmath>#include <algorithm>using namespace std;#define maxn 510000char str[maxn];int find_ans(int k){    if(k<10)    return 1;    int ans=0;    while(k)    {        ans++;        k/=10;    }    return ans;}int find_max(int a,int b,int c,int d){    int ans=0;    int i,j;    for(i=a,j=c;i<b && j<d;i++,j++)    if(str[i]==str[j])    ans++;    else    return ans;    return ans;}int main(){    long long ans1,ans2;    int n,a,b,len1,len2,c,d,num,i,j,k;    while(scanf("%s",str)!=EOF)    {        ans1=ans2=0;        scanf("%d",&n);        c=0,d=0;        for(i=0;i<n;i++)        {            scanf("%d%d",&a,&b);            ans1+=b-a+1;            if(i==0)            {                ans2+=b-a+2+1;            }            else if(c==a)            {                num=min(b-a,d-c);// printf("ans:%d %d\n",num,find_ans(num));                ans2+=(b-a-num+2)+find_ans(num);            }            else            {                num=find_max(a,b,c,d);                ans2+=(b-a-num+2)+find_ans(num);            }            c=a,d=b;        }        prinf("%I64d %I64d\n",ans1,ans2);        //cout<<ans1<<' '<<ans2<<endl;    }    return 0;}