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  • BZOJ 2946 [Poi2000]公共串 (二分+Hash/二分+后缀数组/后缀自动机)

    求多串的最长公共字串.
    法1: 二分长度+hash 传送门
    法2: 二分+后缀数组 传送门
    法3: 后缀自动机
    拿第一个串建自动机,然后用其他串在上面匹配.每次求出SAM上每个节点的最长匹配长度后,再在全局取最小值(因为是所有串的公共串)就行了.

    CODE

    #include<bits/stdc++.h>
    using namespace std;
    char cb[1<<15],*cs=cb,*ct=cb;
    #define getc() (cs==ct&&(ct=(cs=cb)+fread(cb,1,1<<15,stdin),cs==ct)?0:*cs++)
    template<class T>inline void read(T &res) {
    	char ch; int flg = 1; while(!isdigit(ch=getchar()))if(ch=='-')flg=-flg;
    	for(res=ch-'0';isdigit(ch=getchar());res=res*10+ch-'0'); res*=flg;
    }
    const int MAXN = 2005;
    const int C = 26;
    int m, lk[MAXN<<1], ch[MAXN<<1][C], len[MAXN<<1], mx[MAXN<<1], mn[MAXN<<1], bin[MAXN<<1], seq[MAXN<<1], sz, last;
    inline void init() {
        last = sz = 0; ++sz;
        lk[0] = -1; len[0] = 0;
    }
    inline void Copy(int A, int B) {
        lk[A] = lk[B];
        memcpy(ch[A], ch[B], sizeof ch[B]);
    }
    inline int insert(int p, int c) {
        int cur = sz++; last = cur;
        len[cur] = len[p] + 1;
        for(; ~p && !ch[p][c]; p = lk[p]) ch[p][c] = cur;
        if(p == -1) lk[cur] = 0;
        else {
            int q = ch[p][c];
            if(len[p] + 1 == len[q]) lk[cur] = q;
            else {
                int x = sz++;
                len[x] = len[p] + 1;
                Copy(x, q);
                lk[cur] = lk[q] = x;
                for(; ~p && ch[p][c] == q; p = lk[p]) ch[p][c] = x;
            }
        }
        return last;
    }
    inline void Match(char *s) {
        int p = 0, now = 0, j = 0;
        for(int c; s[j]; ++j) {
            if(ch[p][c=s[j]-'a'])
                ++now, p = ch[p][c];
            else {
                for(; ~p && !ch[p][c]; p = lk[p]);
                if(p == -1) now = p = 0;
                else now = len[p] + 1, p = ch[p][c];
            }
            mx[p] = max(mx[p], now);
        }
        for(int i = sz-1; i; --i) {
            mn[seq[i]] = min(mn[seq[i]], mx[seq[i]]);
            mx[lk[seq[i]]] = max(mx[lk[seq[i]]], mx[seq[i]]);
            mx[seq[i]] = 0;
        }
    }
    char s[MAXN];
    int main() {
    	read(m); scanf("%s", s);
    	if(--m == 0) return printf("%d
    ", strlen(s)), 0;
    	init(); int p = 0;
    	for(int i = 0; s[i]; ++i) p = insert(p, s[i]-'a');
        for(int i = 0; i < sz; ++i) ++bin[mn[i] = len[i]];
        for(int i = 1; i < sz; ++i) bin[i] += bin[i-1];
        for(int i = sz-1; ~i; --i) seq[--bin[len[i]]] = i;
    	while(m--) scanf("%s", s), Match(s);
        int ans = 0;
    	for(int i = 1; i < sz; ++i)
            ans = max(ans, mn[i]);
        printf("%d
    ", ans);
    }
    
    
    
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  • 原文地址:https://www.cnblogs.com/Orz-IE/p/12039314.html
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