题目链接:uva 10526 - Intellectual Property
题目大意:给定两个文本,问说下面一个文本中在哪些位置上抄袭了上面个一个文本的,输出n个抄袭位置(不足n个情况全部输出),按照长度优先输出,长度相同的输出位置靠前的。
注意:空格,回车都算一个字符;一段字符只能是抄袭上面的一部分,比如上:NSB*SB 下:NSB 答案:NSB。
解题思路:将两个文本连接在一起,中间用没有出现的字符分割,然后处理处后缀数组,根据height数组的性质,求出哪些位置匹配的长度不为0(注意匹配的位置为后面一段跟前面一段的最大长度),跟据分割符的下标就可以区别前后文本。然后将覆盖的情况处理掉排序输出。
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 200005;
struct state {
int pos, len;
state (int pos = 0, int len = 0) {
this->pos = pos;
this->len = len;
}
};
struct Suffix_Arr {
int n, s[maxn];
int SA[maxn], rank[maxn], height[maxn];
int tmp_one[maxn], tmp_two[maxn], c[maxn];
void init();
void add(char* str);
void build_arr(int m);
void get_height();
void solve (int p);
}AC;
inline bool sort_len (const state& a, const state& b) {
if (a.len != b.len)
return a.len > b.len;
return a.pos < b.pos;
}
inline bool sort_pos (const state& a, const state& b) {
if (a.pos != b.pos)
return a.pos < b.pos;
return a.len > b.len;
}
int N, tick;
char str[maxn];
void init () {
AC.init();
gets(str);
while (gets(str) && strcmp(str, "END TDP CODEBASE"))
AC.add(str);
AC.s[AC.n++] = 260;
tick = AC.n;
gets(str);
while (gets(str) && strcmp(str, "END JCN CODEBASE"))
AC.add(str);
AC.s[AC.n++] = 0;
AC.build_arr(261);
AC.get_height();
}
int main () {
int cas = 0;
while (~scanf("%d%*c", &N) && N) {
init();
if (cas)
printf("\n");
printf("CASE %d\n", ++cas);
AC.solve(tick);
}
return 0;
}
void Suffix_Arr::solve(int p) {
int k = 0;
memset(c, 0, sizeof(c));
for (int i = 0; i < n; i++) {
if (SA[i] < p)
k = height[i+1];
else {
k = min(height[i], k);
c[i] = max(c[i], k);
}
}
k = 0;
for (int i = n - 1; i >= 0; i--) {
if (SA[i] < p) {
k = height[i];
} else {
c[i] = max(c[i], k);
k = min(height[i], k);
}
}
vector<state> vec, ans;
for (int i = 0; i < n; i++) {
if (c[i] >= 1)
vec.push_back(state(SA[i], c[i]));
}
sort(vec.begin(), vec.end(), sort_pos);
int mv = -1;
for (int i = 0; i < vec.size(); i++) {
if (vec[i].pos + vec[i].len <= mv)
continue;
ans.push_back(vec[i]);
mv = vec[i].pos + vec[i].len;
}
sort(ans.begin(), ans.end(), sort_len);
for (int i = 0; i < N && i < ans.size(); i++) {
printf("INFRINGING SEGMENT %d LENGTH %d POSITION %d\n", i+1, ans[i].len, ans[i].pos - p);
for (int j = 0; j < ans[i].len; j++)
printf("%c", s[ans[i].pos + j]);
printf("\n");
}
}
void Suffix_Arr::init() {
n = 0;
memset(s, 0, sizeof(s));
memset(height, 0, sizeof(height));
}
void Suffix_Arr::add (char *str) {
int len = strlen(str);
for (int i = 0; i < len; i++)
s[n++] = str[i];
s[n++] = ‘\n‘;
}
void Suffix_Arr::get_height () {
for (int i = 0; i < n; i++)
rank[SA[i]] = i;
int mv = 0;
for (int i = 0; i < n - 1; i++) {
if (mv) mv--;
int j = SA[rank[i]-1];
while (s[i+mv] == s[j+mv])
mv++;
height[rank[i]] = mv;
}
}
void Suffix_Arr::build_arr(int m) {
int *x = tmp_one, *y = tmp_two;
for (int i = 0; i < m; i++) c[i] = 0;
for (int i = 0; i < n; i++) c[x[i] = s[i]]++;
for (int i = 1; i < m; i++) c[i] += c[i-1];
for (int i = n-1; i >= 0; i--) SA[--c[x[i]]] = i;
for (int k = 1; k <= n; k <<= 1) {
int mv = 0;
for (int i = n - k; i < n; i++) y[mv++] = i;
for (int i = 0; i < n; i++) if (SA[i] >= k)
y[mv++] = SA[i] - k;
for (int i = 0; i < m; i++) c[i] = 0;
for (int i = 0; i < n; i++) c[x[y[i]]]++;
for (int i = 1; i < m; i++) c[i] += c[i-1];
for (int i = n - 1; i >= 0; i--) SA[--c[x[y[i]]]] = y[i];
swap(x, y);
mv = 1;
x[SA[0]] = 0;
for (int i = 1; i < n; i++)
x[SA[i]] = (y[SA[i-1]] == y[SA[i]] && y[SA[i-1] + k] == y[SA[i] + k] ? mv - 1 : mv++);
if (mv >= n)
break;
m = mv;
}
}
时间: 2024-10-11 22:32:12