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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;
	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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原文地址：http://blog.csdn.net/accelerator_/article/details/38776499