cf 427D Match & Catch 后缀数组

将两个字符串连接,中间加一个没出现的字符,最后添加0。然后从小到大枚举长度ans,对于每个ans,对height数组分组,答案只可能出现在某组内,对每组进行检查,如果只存在两个字符串一个在前一个在后则return true,贴的代码是别人的。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100010;
char s[maxn];
int sa[maxn];
int t[maxn], t2[maxn], c[maxn];
int rank[maxn], height[maxn];
int l1, l2;
void build_sa(int m, int 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;
    }
}

void getHeight(int n)
{
    int k = 0;
    for(int i = 0; i <= n; i++)
        rank[sa[i]] = i;
    for(int 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;
    }
}

bool ok(int m, int n)
{
    int cnt1 = 0, cnt2 = 0;
    for(int i = 1; i <= n; i++)
    {
        if(height[i] >= m)
        {
            if(sa[i-1] < l1)
                cnt1++;
            if(sa[i-1] > l1)
                cnt2++;
        }
        else
        {
            if(sa[i-1] < l1)
                cnt1++;
            if(sa[i-1] > l1)
                cnt2++;
            if(cnt1 == 1 && cnt2 == 1)
                return true;
            cnt1 = cnt2 = 0;
        }
    }
    return false;
}
int main()
{
    char a[5555], b[5555];
    while(scanf("%s %s", &a, &b) != EOF)
    {
        l1 = strlen(a);
        l2 = strlen(b);
        int n = 0;
        for(int i = 0; i < l1; i++)
            s[n++] = a[i];
        s[n++] = 'z'+1;
        for(int i = 0; i < l2; i++)
            s[n++] = b[i];
        s[n] = 0;
        build_sa(128, n+1);
        getHeight(n);
        
        int l = 1, r = 5000;
        int ans = -1;
        int len = min(l1, l2);
        for(int i = 1; i <= len; i++)
            if(ok(i, n))
            {
                ans = i;
                break;
            }
        if(ans <= 0)
            printf("-1\n");
        else
            printf("%d\n", ans);
    }
    return 0;
}


 

 

posted @ 2014-05-05 15:30  贝尔摩德  阅读(198)  评论(0编辑  收藏  举报