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UVA 10526 - Intellectual Property
题目链接
题意：给定两个问题，要求找出第二个文本抄袭第一个文本的所有位置和长度，输出前k个，按长度从大到小先排，长度一样的按位置从小到大
思路：后缀数组，把两个文本拼接起来，记录下拼接位置为tdp，这样如果sa[i] < tdp就是前面的文本开头，如果sa[i] >= tdp就是后面的文本开头，拼接起来的求出height数组，利用该数组的性质，从前往后扫一遍，从后往前扫一遍，把所有位置的最大值保存下来，最后在扫描一遍位置，把答案记录下来
代码：
#include <cstdio>
#include <cstring>
#include <algorithm>using namespace std;const int MAXLEN = 200005;
const int INF = 0x3f3f3f3f;char str[55555];
int k, tdp, an, v[MAXLEN];struct Ans {int len, pos;Ans() {}Ans(int len, int pos) {this->len = len;this->pos = pos;}
} ans[MAXLEN];bool cmp(Ans a, Ans b) {if (a.len == b.len) return a.pos < b.pos;return a.len > b.len;
}struct Suffix {int s[MAXLEN];int sa[MAXLEN], t[MAXLEN], t2[MAXLEN], c[MAXLEN], n;int rank[MAXLEN], height[MAXLEN];void build_sa(int m) {n++;int i, *x = t, *y = t2;for (i = 0; i < m; i++) c[i] = 0;for (i = 0; i < n; i++) c[x[i] = s[i]]++;for (i = 1; i < m; i++) c[i] += c[i - 1];for (i = n - 1; i >= 0; i--) sa[--c[x[i]]] = i;for (int k = 1; k <= n; k <<= 1) {int p = 0;for (i = n - k; i < n; i++) y[p++] = i;for (i = 0; i < n; i++) if (sa[i] >= k) y[p++] = sa[i] - k;for (i = 0; i < m; i++) c[i] = 0;for (i = 0; i < n; i++) c[x[y[i]]]++;for (i = 0; i < m; i++) c[i] += c[i - 1];for (i = n - 1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];swap(x, y);p = 1; x[sa[0]] = 0;for (i = 1; i < n; i++)x[sa[i]] = (y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k]) ? p - 1 : p++;if (p >= n) break;m = p;}n--;}void getHeight() {int i, j, k = 0;for (i = 1; i <= n; i++) rank[sa[i]] = i;for (i = 0; i < n; i++) {if (k) k--;int j = sa[rank[i] - 1];while (s[i + k] == s[j + k]) k++;height[rank[i]] = k;}}void init() {tdp = 0; n = 0; an = 0;gets(str);while (gets(str)) {if (strcmp(str, "END TDP CODEBASE") == 0) break;int len = strlen(str);str[len] = '\n';for (int i = 0; i <= len; i++)s[n++] = str[i];}tdp = n;s[n++] = 260;gets(str);while (gets(str)) {if (strcmp(str, "END JCN CODEBASE") == 0) break;int len = strlen(str);str[len] = '\n';for (int i = 0; i <= len; i++)s[n++] = str[i];}s[n] = 0;}void solve() {init();build_sa(261);getHeight();memset(v, 0, sizeof(v));int Min = -1;for (int i = 1; i <= n; i++) {if (sa[i] < tdp) Min = INF;else if (sa[i] > tdp) {if (Min == -1) continue;Min = min(height[i], Min);v[sa[i] - tdp - 1] = max(Min, v[sa[i] - tdp - 1]);}}Min = -1;for (int i = n; i >= 1; i--) {if (sa[i] < tdp) Min = INF;else if (sa[i] > tdp) {if (Min == -1) continue;Min = min(height[i + 1], Min);v[sa[i] - tdp - 1] = max(Min, v[sa[i] - tdp - 1]);}}int r = -1;for (int i = 0; i < n - tdp; i++) {if (i + v[i] <= r) continue;if (v[i] == 0) continue;ans[an++] = Ans(v[i], i);r = i + v[i];}sort(ans, ans + an, cmp);for (int i = 0; i < min(an, k); i++) {printf("INFRINGING SEGMENT %d LENGTH %d POSITION %d\n", i + 1, ans[i].len, ans[i].pos);for (int j = ans[i].pos + tdp + 1; j < ans[i].pos + tdp + 1 + ans[i].len; j++)printf("%c", s[j]);printf("\n");}}} gao;int main() {int bo = 0;int cas = 0;while (~scanf("%d%*c", &k) && k) {if (bo) printf("\n");else bo = 1;printf("CASE %d\n", ++cas);gao.solve();}return 0;
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