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UVA 10526 - Intellectual Property (后缀数组)
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; 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(256); getHeight(); memset(v, 0, sizeof(v)); int Min = -1; for (int i = 1; i <= n; i++) { if (sa[i] < tdp) Min = INF; else { if (Min == -1) continue; Min = min(height[i], Min); v[sa[i] - tdp] = max(Min, v[sa[i] - tdp]); } } Min = -1; for (int i = n; i >= 1; i--) { if (sa[i] < tdp) Min = INF; else { if (Min == -1) continue; Min = min(height[i + 1], Min); v[sa[i] - tdp] = max(Min, v[sa[i] - tdp]); } } 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; j < ans[i].pos + tdp + 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; }
UVA 10526 - Intellectual Property (后缀数组)
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