|hdu 2328|后缀数组|二分|Corporate Identity

hdu 2328

给出n个字符串，输出他们的最长公共子串，无解输出”IDENTITY LOST”

用不同的符号连接每个字符串，然后二分公共子串的长度，在height数组中看有没有连续n个height大于公共子串的长度，如果有，那么更新答案。
(此题暴力比SA快，而且poj上用SA一直TLE，Hdu上1840ms就过了)

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#define ms(i, j) memset(i, j, sizeof i)#define FN2 "poj3450" using namespace std;const int MAXN = 200 + 5, MAXT = 4000 + 5;char s[MAXT][MAXN];int t, n, m, ans[MAXN], ll, mini;int a[MAXT*MAXN+200], SA[MAXT*MAXN+200], rk[MAXT*MAXN+200], tp[MAXT*MAXN+200], tax[MAXT*MAXN+200], height[MAXT*MAXN+200], belong[MAXT*MAXN+200];bool cmp(int *f, int i, int k) {return f[SA[i]]==f[SA[i-1]]&&f[SA[i]+k]==f[SA[i-1]+k];}void build() {    for (int i=0;i<m;i++) tax[i] = 0;    for (int i=0;i<n;i++) tax[rk[i]=a[i]]++;    for (int i=1;i<m;i++) tax[i] += tax[i-1];    for (int i=n-1;i>=0;i--) SA[--tax[rk[i]]] = i;    int p;    for (int k=1;k<=n;k*=2) {        p = 0;        for (int i=n-k;i<n;i++) tp[p++] = i;        for (int i=0;i<n;i++) if (SA[i]>=k) tp[p++] = SA[i] - k;        for (int i=0;i<m;i++) tax[i] = 0;        for (int i=0;i<n;i++) tax[rk[tp[i]]]++;        for (int i=1;i<m;i++) tax[i] += tax[i-1];        for (int i=n-1;i>=0;i--) SA[--tax[rk[tp[i]]]] = tp[i];        swap(rk, tp), p = 0, rk[SA[0]] = 0;        for (int i=1;i<n;i++) rk[SA[i]] = cmp(tp, i, k) ? p : ++p;        if (++p>=n) break;        m = p;    }}void getH() {    int k = 0;    for (int i=0;i<n;i++) {        if (k) k--;        int j = SA[rk[i]-1];        while (a[i+k]==a[j+k]) k++;        height[rk[i]] = k;    }}void init() {    n = 0;    int orz = 200;    mini = 200000000;    for (int i=0;i<t;i++) {        scanf("%s", s[i]);        int l = strlen(s[i]); mini = min(mini, l);//取mini优化二分次数         for (int j=0;j<l;j++) belong[n] = i, a[n++] = s[i][j];        a[n] = orz;        a[n++] = orz++;    }       a[n-1] = 0;    m = orz + 5;}int used[MAXT];bool check(int x) {    int tot = 0;    ms(used, false);    for (int i=2;i<n;i++) {        if (height[i]<x) {            tot = 0;            ms(used, false);            continue;        }        if (!used[belong[SA[i-1]]]) used[belong[SA[i-1]]] = true, ++tot;        if (!used[belong[SA[i]]]) used[belong[SA[i]]] = true, ++tot;        if (tot>=t) {            int k = 0;            for (int j=SA[i-1];j<SA[i-1]+height[i-1];j++) ans[k++] = a[j];            return true;        }     }    return false;}void solve() {    build(), getH();    int l = 0, r = mini+1; ll = 0;//左闭右开，要+1！ /*  for (int i=0;i<n;i++) {        for (int j=SA[i];j<n;j++) printf("%4d ", a[j]);        putchar('\n');    }    return ;*/    while (l<r) {        int mid = (l+r)/2;        if (check(mid)) {            ll = mid;            l = mid + 1;        } else r = mid;    }    if (ll==0) printf("IDENTITY LOST\n");     else {        for (int i=0;i<ll;i++) printf("%c", ans[i]);        printf("\n");    }}int main() {    #ifndef ONLINE_JUDGE    freopen(FN2".in","r",stdin);freopen("1.out","w",stdout);    #endif    while(scanf("%d", &t)==1&t) init(), solve();    return 0;}

附上比SA快的暴力(可以对拍)

#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#define ms(i, j) memset(i, j, sizeof i)#define FN2 "poj3450" using namespace std;const int MAXN = 200 + 5, MAXT = 4000 + 5;char s[MAXT][MAXN], ch[MAXN];int t, n;void init() {    for (int i=0;i<t;i++) {        scanf("%s", s[i]);    }       n = strlen(s[0]);}void solve() {    int ans = 0;    char fi[MAXN];    for (int bg=0;bg<n;bg++) {        for (int end=bg;end<n;end++) {            int k = 0;            int siz = end - bg + 1;            if (siz<ans) continue;            for (int i=0;i<n;i++) ch[i] = 0;             for (int i=bg;i<=end;i++) ch[k++] = s[0][i];            bool flag = false;            for (int i=1;i<t;i++) if (!strstr(s[i], ch)) {flag = true; break;}            if (!flag) {                if (siz==ans&&strcmp(ch, fi)>=0) continue;                memcpy(fi, ch, sizeof ch), ans = siz;            }        }    }    if (ans==0) printf("IDENTITY LOST\n");    else printf("%s\n", fi);}int main() {    #ifndef ONLINE_JUDGE    freopen(FN2".in","r",stdin);freopen("2.out","w",stdout);    #endif    while(scanf("%d", &t)==1&&t) init(), solve();    return 0;}