首页 > 代码库 > POJ 2774 Long Long Message 后缀数组

POJ 2774 Long Long Message 后缀数组

最长公共子串问题的后缀数组解法。

将第二个字符串拼接到第一个字符串中去,中间用分割符隔开,避免后面计算的时候发生一个字串横跨两个字符串的情况。

之后看相邻的不在同一个字符串中的lcp的最大值即可。

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <climits>#include <string>using namespace std; #define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int, int> PII;typedef pair<double, double> PDD;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 4e5 + 10;//以下是倍增法求后缀数组int wa[maxn], wb[maxn], wv[maxn], ws[maxn];int cmp(int *r, int a, int b, int l) { return r[a] == r[b] && r[a + l] == r[b + l]; }void da(int *r, int *sa, int n, int m) {    int i, j, p, *x = wa, *y = wb, *t;    for(i = 0; i < m; i++) ws[i] = 0;    for(i = 0; i < n; i++) ws[x[i] = r[i]]++;    for(i = 1; i < m; i++) ws[i] += ws[i - 1];    for(i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;    for(j = 1, p = 1; p < n; j <<= 1, m = p) {        for(p = 0, i = n - j; i < n; i++) y[p++] = i;        for(i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] - j;        for(i = 0; i < n; i++) wv[i] = x[y[i]];        for(i = 0; i < m; i++) ws[i] = 0;        for(i = 0; i < n; i++) ws[wv[i]]++;        for(i = 0; i < m; i++) ws[i] += ws[i - 1];        for(i = n - 1; i >= 0; i--) sa[--ws[wv[i]]] = y[i];        for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)             x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;    }}//以下是求解height 数组int height[maxn], Rank[maxn];void calheight(int *r, int *sa, int n) {    int i, j, k = 0;    for(i = 1; i <= n; i++) Rank[sa[i]] = i;    for(i = 0; i < n; height[Rank[i++]] = k)         for(k ? k-- : 0, j = sa[Rank[i] - 1];                 r[i + k] == r[j + k]; k++) ;}char buf[maxn];int str[maxn], sa[maxn], len, n, len1;int main() {    while(scanf("%s", buf) != EOF) {        len1 = len = strlen(buf);        for(int i = 0; i < len; i++) str[n++] = buf[i] + 1;        str[n++] = ‘$‘ + 1;        scanf("%s", buf);        len = strlen(buf);        for(int i = 0; i < len; i++) str[n++] = buf[i] + 1;        buf[n] = 0;        da(str, sa, n + 1, 200);        calheight(str, sa, n);        int ans = 0;        for(int i = 1; i <= n; i++) if((sa[i] < len1) ^ (sa[i - 1] < len1)) {            ans = max(ans, height[i]);        }        printf("%d\n", ans);    }    return 0;}

 

POJ 2774 Long Long Message 后缀数组