Codeforces 427D Match & Catch 后缀自动机 或 后缀数组

题目大意：

就是对于两个字符串S1和S2, 求出他们最短的一个公共字串P, 满足P在S1中只出现1次, P在S2中也只出现一次, 输出P最小的长度, 如果这样的P不存在, 输出-1

大致思路：

很明显的一个做法是将S1和S2连接起来中间用未出现的字符隔开, 然后建立后缀自动机, 记录每一个状态中表示的字符串的来源(可以状压记录), 然后所有状态中, 满足Right集合为2, 且状压表示出现在了2个字符串中即可, 在这样一个状态中取最短可行长度, 遍历所有状态点即可, 复杂度O(|S1| + |S2|)

当然还可以用后缀数组来做, 考虑到P要满足的条件如果这样的P存在, 必定是sa数组中相邻的存在, 所以扫描一遍height数组, 当sa[i - 1]与sa[i]不属于同一字符串时, 考虑P不能再其他位置出现那么只能height[i - 1] < height[i], height[i + 1] < height[i]是存在这样的P, 最小长度为 max(height[i - 1], height[i + 1]) + 1, 扫描一遍height数组找出满足条件的最小的那一个P即可

后缀自动机代码如下：

Result  :  Accepted     Memory  :  2648 KB     Time  :  31 ms

/* * Author: Gatevin * Created Time:  2015/4/16 19:29:08 * File Name: Rin_Tohsaka.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endl#define maxm 10010#define maxn 20010struct Suffix_Automation{    struct State    {        State *par;        State *go[27];        int right, val, mi, appear;        void init(int _val = 0)        {            par = 0, val = _val, right = mi = appear = 0;            memset(go, 0, sizeof(go));        }    };    State *root, *last, *cur;    State nodePool[maxn];    State* newState(int val = 0)    {        cur->init(val);        return cur++;    }    void initSAM()    {        cur = nodePool;        root = newState();        last = root;    }    void extend(int w, int belong)    {        State *p = last;        State *np = newState(p->val + 1);        np->right = 1;        np->appear = (1 << belong);        while(p && p->go[w] == 0)        {            p->go[w] = np;            p = p->par;        }        if(p == 0)        {            np->par = root;        }        else        {            State *q = p->go[w];            if(q->val == p->val + 1)                np->par = q;            else            {                State *nq = newState(p->val + 1);                memcpy(nq->go, q->go, sizeof(q->go));                nq->par = q->par;                q->par = nq;                np->par = nq;                while(p && p->go[w] == q)                {                    p->go[w] = nq;                    p = p->par;                }            }        }        last = np;    }    int d[maxm];    State *b[maxn];    void topo()    {        int cnt = cur - nodePool;        int maxVal = 0;        memset(d, 0, sizeof(d));        for(int i = 1; i < cnt; i++)            maxVal = max(maxVal, nodePool[i].val), d[nodePool[i].val]++;        for(int i = 1; i <= maxVal; i++) d[i] += d[i - 1];        for(int i = 1; i < cnt; i++) b[d[nodePool[i].val]--] = &nodePool[i];        b[0] = root;    }    void SAMInfo()    {        State *p;        int cnt = cur - nodePool;        for(int i = cnt - 1; i > 0; i--)        {            p = b[i];            p->par->right += p->right;            p->par->appear |= p->appear;            p->mi = p->par->val + 1;        }    }};Suffix_Automation sam;char s[maxm];int main(){    scanf("%s", s);    int len = strlen(s);    sam.initSAM();    for(int i = 0; i < len; i++)        sam.extend(s[i] - 'a', 0);    sam.extend(26, 2);    scanf("%s", s);    len = strlen(s);    for(int i = 0; i < len; i++)        sam.extend(s[i] - 'a', 1);    sam.topo();    sam.SAMInfo();    int ans = 1e9;    int cnt = sam.cur - sam.nodePool;    for(int i = cnt - 1; i > 0; i--)        if(sam.b[i]->appear == (1 << 2) - 1 && sam.b[i]->right == 2)//这个串只出现两次且属于不同串            ans = min(ans, sam.b[i]->mi);//取长度最小的即可    if(ans == 1e9) ans = -1;    printf("%d\n", ans);    return 0;}

后缀数组做法如下：

Result  :  Accepted     Memory  :  400  KB     Time  :  31 ms

/* * Author: Gatevin * Created Time:  2015/4/16 20:11:25 * File Name: Rin_Tohsaka.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endl#define maxn 10010int 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(int *r, int *sa, int n, int m){    int *x = wa, *y = wb, *t, i, j, p;    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 <<= 1, 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(int *r, int *sa, int n){    int i, j, k = 0;    for(i = 1; i <= n; i++) rank[sa[i]] = i;    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 in[maxn];int s[maxn], sa[maxn], belong[maxn];int N;int main(){    scanf("%s", in);    int len = strlen(in);    N = 0;    for(int i = 0; i < len; i++)        s[N] = in[i] - 'a' + 1, belong[N++] = 1;    belong[N] = 0;    s[N++] = 27;//刚开始逗比地写的26就WA了...    scanf("%s", in);    len = strlen(in);    for(int i = 0; i < len; i++)        s[N] = in[i] - 'a' + 1, belong[N++] = 2;    s[N] = 0;    da(s, sa, N + 1, 30);    calheight(s, sa, N);    int ans = 1e9;    for(int i = 1; i <= N; i++) if(height[i] > 0)//s[sa[0]] = 0, s[sa[N]] = 26不考虑        if(belong[sa[i - 1]] != belong[sa[i]])//说明这两个不属于同一个串, 接下来保证P在各个串中只出现一次        {            if(i == N && height[i - 1] < height[i])                ans = min(ans, height[i - 1] + 1);            if(i == 1 && height[i + 1] < height[i])                ans = min(ans, height[i + 1] + 1);            if(i > 1 && i < N && height[i - 1] < height[i] && height[i + 1] < height[i])                ans = min(ans, max(height[i - 1], height[i + 1]) + 1);        }    if(ans == 1e9) ans = -1;    printf("%d\n", ans);    return 0;}