POJ 3415 Common Substrings 后缀数组

题目：

http://poj.org/problem?id=3415

题意：

给定两个字符串，求这两个字符串中长度大于等于k的公共子串的数量

思路：

不得不说，这题我不会，看的别人的，心累。。。看这篇http://www.cnblogs.com/luxiaoming/p/5270984.html

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;
const int N = 200000 + 10, INF = 0x3f3f3f3f;

int sa[N], height[N], rnk[N], wa[N], wb[N], c[N];
char str[N];
int s[N];
int stk[N][2];

bool cmp(int *r, int a, int b, int l)
{
    return r[a] == r[b] && r[a+l] == r[b+l];
}
void Rsort(int *x, int *y, int n, int m)
{
    for(int i = 0; i < m; i++) c[i] = 0;
    for(int i = 0; i < n; i++) c[x[y[i]]]++;
    for(int i = 1; i < m; i++) c[i] += c[i-1];
    for(int i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];
}
void da(int *s, int n, int m)
{
    int *x = wa, *y = wb;
    for(int i = 0; i < n; i++) x[i] = s[i], y[i] = i;
    Rsort(x, y, n, m);
    for(int j = 1, p = 1; p < n; j *= 2, m = p)
    {
        p = 0;
        for(int i = n-j; i < n; i++) y[p++] = i;
        for(int i = 0; i < n; i++) if(sa[i] >= j) y[p++] = sa[i] - j;
        Rsort(x, y, n, m);
        swap(x, y); p = 1; x[sa[0]] = 0;
        for(int i = 1; i < n; i++) x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p-1 : p++;
    }
}
void get_height(int *s, int n)
{
    int i, j, k = 0;
    for(i = 0; i <= n; i++) rnk[sa[i]] = i;
    for(i = 0; i < n; height[rnk[i++]] = k)
        for(k ? --k : 0, j = sa[rnk[i]-1]; s[i+k] == s[j+k]; k++);
}
int main()
{
    int k;
    while(scanf("%d", &k), k)
    {
        scanf("%s", str);
        int len = 0, len_1 = strlen(str);
        for(int i = 0; str[i]; i++) s[len++] = str[i];
        s[len++] = '$';
        scanf("%s", str);
        for(int i = 0; str[i]; i++) s[len++] = str[i];
        s[len] = 0;
        da(s, len+1, 150);
        get_height(s, len);
        ll ans = 0, num = 0;
        int top = 0;
        for(int i = 2; i <= len; i++)
        {
            if(height[i] < k) top = 0, num = 0;
            else
            {
                int tmp = 0;
                if(sa[i-1] < len_1) tmp++, num += height[i] - k + 1;
                while(top > 0 && height[i] <= stk[top-1][0])
                {
                    top--;
                    num -= stk[top][1] * (stk[top][0] - height[i]);
                    tmp += stk[top][1];
                }
                stk[top][0] = height[i], stk[top++][1] = tmp;
                if(sa[i] > len_1) ans += num;
            }
        }
        top = 0, num = 0;
        for(int i = 2; i <= len; i++)
        {
            if(height[i] < k) top = 0, num = 0;
            else
            {
                int tmp = 0;
                if(sa[i-1] > len_1) tmp++, num += height[i] - k + 1;
                while(top > 0 && height[i] <= stk[top-1][0])
                {
                    top--;
                    num -= stk[top][1] * (stk[top][0] - height[i]);
                    tmp += stk[top][1];
                }
                stk[top][0] = height[i], stk[top++][1] = tmp;
                if(sa[i] < len_1) ans += num;
            }
        }
//        for(int i = 0; i < len_1; i++)
//            for(int j = len_1 + 1; j < len; j++)
//            {
//                int st = min(rnk[i], rnk[j]), en = max(rnk[i], rnk[j]);
//                int t = INF;
//                for(int q = st+1; q <= en; q++) t = min(t, height[q]);//这步可以优化掉，不过无所谓啦。。。
//                if(t >= k)ans += t - k + 1;
//            }
        printf("%lld\n", ans);
    }
    return 0;
}