URAL 1517 Freedom of Choice 后缀数组

来源：互联网发布：轴类零件数控加工编程编辑：程序博客网时间：2024/06/05 16:11

题目大意：

就是对于给出两个长度都为N的只包含大写字母的串求其最长公共子串

存在多个长度相同的的串时输出任意一个

大致思路：

就是一个简单的后缀数组的题...很简单了...

细节见代码

代码如下：

Result : Accepted Memory : 6774 KB Time : 218 ms

/* * Author: Gatevin * Created Time:  2015/2/9 16:05:51 * File Name: Iris_Freyja.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define maxn 233333/* * Doubling Algorithm 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 *x = wa, *y = wb, *t, i, j, p;    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 *= 2, 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 = 1; 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++;    }    return;}int rank[maxn], height[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++);    return;}int N;int s[maxn], sa[maxn];char tmp[maxn >> 1];int start;bool check(int mid, int n){    for(int i = 1; i <= n; i++)        if(height[i] >= mid)        {            if(((lint)sa[i] - N)*((lint)sa[i - 1] - N) < 0)//如果sa[i]与sa[i - 1]来自不同的串            {                start = sa[i];                return true;            }        }    return false;}int main(){    scanf("%d", &N);    scanf("%s", tmp);    for(int i = 0; i < N; i++)        s[i] = tmp[i] - 'A' + 1;    s[N] = 27;    scanf("%s", tmp);    for(int i = 0; i < N; i++)        s[i + 1 + N] = tmp[i] - 'A' + 1;    s[2*N + 1] = 0;    da(s, sa, 2*N + 1 + 1, 30);    calheight(s, sa, 2*N + 1);    int L = 1, R = N, mid, ans = 0;    //二分可能的长度, 因为如果长度len可行那么比len小的都可以,具有单调性    while(L <= R)    {        mid = (L + R) >> 1;        if(check(mid, 2*N + 1))        {            ans = mid;            L = mid + 1;        }        else            R = mid - 1;    }    for(int i = 0; i < ans; i++)        printf("%c", s[start + i] + 'A' - 1);    return 0;}

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