poj 3294(后缀数组)

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
, wb
, ws
, wv
, sa
;
int rank
, height
, s
, n, cnt, sum[115], vis[115], pos
;

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
;
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;
}