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UVA 11107 - Life Forms

题目链接

题意：给定一些字符串，求最长并且在所有字符串的连续子串中出现超过一半次数的字符串，输出这些字符串

思路：把这些字符串接起来，拼接部分用一个不会出现的不重复的字符，然后求这个长串的后缀数组，利用height数组去进行二分求解，二分的判断里面如果有一个连续height段超过了一半次数，那么就是可行的，如果所有连续段都没有出现超过一半，就是不可行的

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;

const int MAXLEN = 200005;

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) {
	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;
	}	    
    }

    void getHeight() {
	int i, j, k = 0;
	for (i = 0; i < n; i++) rank[sa[i]] = i;
	for (i = 0; i < n; i++) {
	    if (k) k--;
	    if (rank[i] == 0) continue;
	    int j = sa[rank[i] - 1];
	    while (s[i + k] == s[j + k]) k++;
	    height[rank[i]] = k;
	}
    }
} gao;

const int N = 1005;

int n, l, r, id[MAXLEN];
char str[N];

bool judge(int x, int bo) {
    set<int> vis;    
    vis.insert(id[gao.sa[1]]);
    for (int i = 2; i < gao.n; i++) {
	while (i < gao.n && gao.height[i] >= x) {
	    vis.insert(id[gao.sa[i]]);
	    i++;
	}
	if (vis.size() * 2 > n) {
	    if (bo == 0)
		return true;
	    for (int j = 0; j < x; j++)
		printf("%c", gao.s[gao.sa[i - 1] + j]);
	    printf("\n");
	}
	vis.clear();
	vis.insert(id[gao.sa[i]]);
    }
    return false;
}

void solve() {
    if (!judge(1, 0)) {
	printf("?\n");
	return;
    }
    l = 1; r++;
    while (l < r) {
	int mid = (l + r) / 2;
	if (judge(mid, 0)) l = mid + 1;
	else r = mid;
    }
    l--;
    judge(l, 1);
}

int main() {
    int bo = 0;
    while (~scanf("%d", &n) && n) {
	if (bo) printf("\n");
	else bo = 1;
	if (n == 1) {
	    scanf("%s", str);
	    printf("%s\n", str);
	    continue;
	}
	int tot = 0;
	r = 0;
	for (int i = 0; i < n; i++) {
	    scanf("%s", str);
	    int len = strlen(str);
	    r = max(len, r);
	    for (int j = 0; j < len; j++) {
		id[tot] = i;
		gao.s[tot++] = str[j];
	    }
	    id[tot] = i;
	    gao.s[tot++] = 'z' + i + 1;
	}
	gao.n = tot;
	gao.build_sa('z' + n + 1);
	gao.getHeight();
	solve();
    }
    return 0;
}

UVA 11107 - Life Forms(后缀数组),布布扣,bubuko.com

UVA 11107 - Life Forms(后缀数组)

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原文地址：http://blog.csdn.net/accelerator_/article/details/38662635