SPOJ

SPOJ - NSUBSTR

给定字符串S，求长度为K的子串，最多出现多少次。

后缀自动机+拓扑排序+DP

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>#include <bitset>#define INF 0x3f3f3f3f#define eps 1e-6#define PI 3.1415926#define mod 1000000009#define base 2333using namespace std;typedef long long LL;const int maxn = 3e5 + 10;const int maxx = 1e3 + 10;inline void splay(int &v) {    v=0;char c=0;int p=1;    while(c<'0' || c >'9'){if(c=='-')p=-1;c=getchar();}    while(c>='0' && c<='9'){v=(v<<3)+(v<<1)+c-'0';c=getchar();}    v*=p;}struct Node {    Node *pre, *nxt[26];    int step, v;    void Clear() {        step = 0, v = 0;        pre = NULL;        memset(nxt, NULL, sizeof(nxt));    }} *root, *last;Node st[maxn<<1], *top[maxn<<1], *cur;char str[maxn];int cnt[maxn], dp[maxn];void init() {    cur = st;    root = last = cur++;    root->Clear();}void extend(int w) {    Node *np = cur++, *p = last;    np->Clear();    np->step = p->step+1;    while(p && !p->nxt[w])        p->nxt[w] = np, p = p->pre;    if(p == NULL)        np->pre = root;    else {        Node *q = p->nxt[w];        if(q->step == p->step+1)            np->pre = q;        else {            Node *nq = cur++;            nq->Clear();            memcpy(nq->nxt, q->nxt, sizeof(q->nxt));            nq->step = p->step+1;            nq->pre = q->pre;            np->pre = q->pre = nq;            while(p && p->nxt[w] == q)                p->nxt[w] = nq, p = p->pre;        }    }    last = np;}void solve() {    while(scanf("%s", str) != EOF) {        init();        int len = strlen(str);        for(int i = 0; i < len; i++)            extend(str[i]-'a');        memset(cnt, 0, sizeof(cnt));        memset(dp, 0, sizeof(dp));        for(Node *p = st; p != cur; p++)            cnt[p->step]++;        for(int i = 1; i <= len; i++)            cnt[i] += cnt[i-1];        for(Node *p = st; p != cur; p++)            top[--cnt[p->step]] = p;        for(int i = 0; i < len; i++) {            root = root->nxt[str[i]-'a'];            root->v = 1;        }        int num = cur-st;        for(int i = num-1; i > 0; i--) {            dp[top[i]->step] = max(dp[top[i]->step], top[i]->v);            if(top[i]->pre) top[i]->pre->v += top[i]->v;        }        for(int i = len-1; i >= 0; i--)            dp[i] = max(dp[i], dp[i+1]);        for(int i = 1; i <= len; i++)            printf("%d\n", dp[i]);    }}int main() {    //srand(time(NULL));    //freopen("kingdom.in","r",stdin);    //freopen("kingdom.out","w",stdout);    solve();}