poj1743(后缀数组)

转自：http://www.cnblogs.com/ziyi--caolu/p/3195342.html

http://poj.org/problem?id=1743

题意：给出一串字符，求不重合的最长重复子串..........

我自己的一点想法：编完后发现，其实就是将height值分组，然后记录在二分答案时满足height值>=p的sa[i]的最大最小值，然后要是最大值减去最小值会>=p，这就说明两个子串的lcp值>=p并且它们的坐标也相差>=p，就自然满足题意.........

#include<iostream>
#include<algorithm>
#include<string>
#include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};
#include<set>//int gcd(int a,int b){return b?gcd(b,a%b):a;}
#include<vector>
#include<cmath>
#include<queue>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define mod 1e9+7
#define ll long long
using namespace std;
#define maxn 20050
int wa[maxn], wb[maxn], wv[maxn], wc[maxn];
int r[maxn], sa[maxn], rank[maxn], height[maxn];
int n;

int cmp(int *r, int a, int b, int l) {
   return r[a] == r[b] && r[a+l] == r[b+l];
}

void da() {
   //m为最大字符
   int i, j, p, *x = wa, *y = wb, *t, m = 256;
   for(i = 0; i < m; i++) 	wc[i] = 0;
   for(i = 0; i <= n; i++) 	wc[x[i] = r[i]]++;
   for(i = 1; i < m; i++) 	wc[i] += wc[i-1];
   for(i = n; i >= 0; i--) 	sa[--wc[x[i]]] = i;
   for(j = 1, p = 1; p < n; j *= 2, m = p) {
       for(p = 0, i = n - j + 1; 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++) 	wc[i] = 0;
       for(i = 0; i <= n; i++) 	wc[wv[i]]++;
       for(i = 1; i < m; i++) 	wc[i] += wc[i-1];
       for(i = n; i >= 0; i--) 	sa[--wc[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++;
   }
}

void calheight() {
   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++);
}

bool check(int mid) {
   int minn = sa[1], maxx = sa[1];
   for(int i = 2; i <= n; i++) {
       if(height[i] >= mid) {
           minn = min(minn, sa[i]);
           maxx = max(maxx, sa[i]);
           if(maxx - minn >= mid) return 1;
       } else minn = maxx = sa[i];
   }
   return false;
}

void solve() {
   da();//求sa数组
   calheight();//求rank数组和height数组
   int l = 1, r = n, ans = -1;
   while(l <= r) {
       int mid = (l + r) >> 1;
       if(check(mid)) ans = mid, l = mid + 1;
       else r = mid - 1;
   }
   if(ans >= 4) printf("%d\n", ans + 1);
   else printf("0\n");
}

int main() {
   int a, b;
   while(~scanf("%d", &n)) {
       if(!n) break;
       n--;
       scanf("%d", &b);
       for(int i = 0; i < n; i++) {
           scanf("%d", &a);
           r[i] = a - b + 100;
           b = a;
       }
       r[n] = 0;
       solve();
   }
   return 0;
}

原文地址：poj 3261 : Milk Patterns (后缀数组) 作者：依然

题意：可重叠的k次最长重复子串。

思路：后缀数组。先二分答案，然后将后缀分成若干组。这里要判断的是有没有一个组的后缀个数不小于k。如果有，那么存在k个相同的子串满足条件，否则不存在。时间复杂度为O(nlogn)。 

#include<iostream>
using namespace std;
const int MAX = 20050;
const int M = 20000;    //  M的值题意应为1000000，则对应的wd[]的长度也应为1000000。

int n, k, num[MAX];
int sa[MAX], rank[MAX], height[MAX];
int wa[MAX], wb[MAX], wv[MAX], wd[MAX];

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 n, int m)
{
    int i, j, p, *x = wa, *y = wb, *t;
    for(i = 0; i < m; i ++) wd[i] = 0;
    for(i = 0; i < n; i ++) wd[x[i]=r[i]] ++;
    for(i = 1; i < m; i ++) wd[i] += wd[i-1];
    for(i = n-1; i >= 0; i --) sa[-- wd[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 ++) wd[i] = 0;
        for(i = 0; i < n; i ++) wd[wv[i]] ++;
        for(i = 1; i < m; i ++) wd[i] += wd[i-1];
        for(i = n-1; i >= 0; i --) sa[-- wd[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 ++;
        }
    }
}

void calHeight(int *r, 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 ++);
    }
}

bool valid(int len)
{
    int i = 2, cnt;
    while(1)
    {
        while(i <= n && height[i] < len) i ++;
        if(i > n) break;
        cnt = 1;
        while(i <= n && height[i] >= len)
        {
            cnt ++;
            i ++;
        }
        if(cnt >= k) return true;
    }
    return false;
}

int main()
{
    cin >> n >> k;
    for(int i = 0; i < n; i ++)
    {
        cin >> num[i]; 
    }
    num[n] = 0;
    da(num, n + 1, M);
    calHeight(num, n);

    int low = 1, high = n, mid;
    while(low < high)
    {
        mid = (low+high+1) / 2;
        if(valid(mid)) {
            low = mid;
        }else {
            high = mid - 1;
        }
    }
    cout << low << endl;
    return 0;
}