poj 3294(后缀数组)

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题意：给出n个字符串，求出现次数超过n/2次及以上的最长子串，如果有多组按字典序输出，没有输出?。
题解：多个字符串的处理方式是把所有串连接起来，用没有出现在串中且字典序最大的字符分割，然后二分出一个长度x去判断是否有长度不小于x的公共前缀出现了n/2次以上。需要注意为了判断两个公共前缀不在一个串中，要把每个串的起始位置记录到sum数组，然后连续的height[i]>=x情况下，如果sa[i]所在位置对应的串是未被访问过的就标记已访问且出现次数加1，sa[i-1]同理。如果总次数大于n/2说明有解。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 201000;char str[1005];int wa[N], wb[N], ws[N], wv[N], sa[N];int rank[N], height[N], s[N], n, cnt, sum[115], vis[115], pos[N];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 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 = 0; 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++;    }}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++);}int judge(int x) {    memset(vis, 0, sizeof(vis));    int flag = 0, temp = 0;    for (int i = 1; i <= cnt; i++) {        if (height[i] >= x) {            for (int j = 1; j <= n; j++) {                if (sa[i] > sum[j - 1] && sa[i] < sum[j]) {                    if (!vis[j]) temp++;                    vis[j] = 1;                 }                if (sa[i - 1] > sum[j - 1] && sa[i - 1] < sum[j]) {                    if (!vis[j]) temp++;                    vis[j] = 1;                }            }        }        else {            if (temp > n / 2) pos[++flag] = sa[i - 1];            temp = 0;            memset(vis, 0, sizeof(vis));        }    }    if (temp > n / 2) pos[++flag] = sa[n];    return flag;}int main() {    int t = 0;    while (scanf("%d", &n) == 1 && n) {        int maxx = -1;        cnt = 0;        for (int i = 0; i < n; i++) {            scanf("%s", str);            int len = strlen(str);            maxx = max(maxx, len);            for (int j = 0; j < len; j++)                s[cnt++] = str[j] - 'a' + 1;            s[sum[i + 1] = cnt++] = i + 100;        }        s[cnt - 1] = 0;        DA(s, sa, cnt, 200);        calheight(s, sa, cnt - 1);        int l = 0, r = maxx + 5;        int num = 0;        while (l < r) {            int mid = (l + r + 1) / 2;            int temp = judge(mid);            if (temp) {                l = mid;                num = temp;            }            else                r = mid - 1;        }        if (t++) printf("\n");        if (!l) printf("?\n");        else {            for (int i = 1; i <= num; i++) {                for (int j = pos[i]; j < pos[i] + l; j++)                    printf("%c", s[j] + 'a' - 1);                printf("\n");            }        }    }    return 0;}

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