POJ 3261 字符串

题意：给定一个字符串，求最少出现K次的最长重复字串，这K个字串可以重叠。

做法：1.hash（二分最大长度，这里要判断的是枚举起点，然后对该长度的字串求hash值，统计下该字串出现了几次，如果>=k，则满足条件）

    2.后缀数组（论文经典题，详细看论文）

code：

1.

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <vector>#include <queue>#include <map>#include <set>#include <cmath>#include <string>#define zero(x) (((x)>0?(x):-(x))<eps)#define MAGIC 1121117#define eps 1e-8#define ULL unsigned long long#define Test puts("END")using namespace std;const int MOD = 1000000007;const int INF = 1000000000;const int N = 100005;const int M = 311117;int a[N],n,K,hash[N],fac[N];int times[M],value[M];bool used[M];void makeHash(){    hash[0] = a[0];    for(int i = 1;i < n;i ++){        hash[i] = hash[i - 1] * MAGIC + a[i];    }}int getPos(int v){    int pos = (v % M + M) % M;    while(used[pos] && value[pos] != v){        pos ++;        // cout << "pos " << pos << endl;        if(pos >= M){            pos -= M;        }    }    return pos;}int getHash(int l,int r){    if(l == 0) return hash[r];    else return hash[r] - hash[l - 1] * fac[r - l + 1];}int insert(int v){    int pos = getPos(v);    if(!used[pos]){        used[pos] = true;        value[pos] = v;    }    times[pos] ++;    return times[pos];}bool check(int limit){    memset(times,0,sizeof(times));    memset(used,false,sizeof(used));    for(int l = 0;l + limit - 1 < n;l ++){        int r = l + limit - 1;        int v = getHash(l,r);        int ret = insert(v);        if(ret >= K) return true;    }    return false;}int main(){    // freopen("input.txt","r",stdin);    while(scanf("%d%d",&n,&K) != EOF){        for(int i = 0;i < n;i ++){            scanf("%d",&a[i]);        }        fac[0] = 1;        for(int i = 1;i < N;i ++)            fac[i] = fac[i - 1] * MAGIC;        makeHash();        int l = 0,r = n;        int ans = -1;        while(l <= r){            int mid = (l + r) >> 1;            // cout << l << ' ' << r << endl;            if(check(mid)){                ans = mid;                l = mid + 1;            }            else r = mid - 1;        }        printf("%d\n",ans);    }    return 0;}

2.后缀数组

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <vector>#include <queue>#include <map>#include <set>#include <cmath>#include <string>#define zero(x) (((x)>0?(x):-(x))<eps)#define MAGIC 11117#define eps 1e-8#define LL long long#define Test puts("END")using namespace std;const int MOD = 1000000007;const int INF = 1000000000;const int N = 2000100;const int M = 2000100;int s[N],n,K;int wa[N],wb[N],wv[N],wc[N],rank[N],height[N],sa[N];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 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 ++) 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 - 1;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;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 - 1;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 ++;    }    return ;}bool check(int limit){    int l = 0;    while(l <= n){        if(height[l] < limit){            l ++;            continue;        }        int r = l;        while(r <= n && height[r] >= limit){            r ++;        }        if(r - l + 1>= K) return true;        l = r;    }    return false;}int main(){    // freopen("input.txt","r",stdin);    while(scanf("%d%d",&n,&K) != EOF){        for(int i = 0;i < n;i ++){            scanf("%d",&s[i]);        }        da(s,sa,n + 1,200);        calheight(s,sa,n);        /*for(int i = 0;i <= n;i ++)            printf("i:%d  sa:%d  height:%d\n",i,sa[i],height[i]);*/        // check(4);        int l = 0,r = n;        int ans = -1;        while(l <= r){            int mid = (l + r) >> 1;            // cout << l  << ' ' << r << endl;            if(check(mid)){                ans = mid;                l = mid + 1;            }            else r = mid - 1;        }        printf("%d\n",ans);    }    return 0;}