poj 1743(二分+后缀数组)

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poj 1743

注意用$height$分组如果最后一组最后一个元素在序列末，那么要进行处理！最方便是直接$<=n$，还要注意的是不能重复，而且是$mini+x<maxi$不能是$mini+x<=maxi$

本题求的是长度最少为5的重复子串，并且重复子串可以加上或者减去一个数。我们将数字处理一下取差值，然后直接做，之后再结果$+1$即可

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define ms(i, j) memset(i, j, sizeof i)
using namespace std;

const int MAXN = 20000 + 5;

int n, m, a[MAXN], SA[MAXN], tp[MAXN], rk[MAXN], tax[MAXN], height[MAXN];

bool cmp(int *f, int i, int k) {return f[SA[i]]==f[SA[i-1]]&&f[SA[i]+k]==f[SA[i-1]+k];}
void build_SA() {
    for (int i=0;i<m;i++) tax[i] = 0;
    for (int i=0;i<n;i++) tax[rk[i]=a[i]]++;
    for (int i=1;i<m;i++) tax[i] += tax[i-1];
    for (int i=n-1;i>=0;i--) SA[--tax[rk[i]]] = i;
    int p;
    for (int k=1;k<=n;k*=2) {
        p = 0;
        for (int i=n-k;i<n;i++) tp[p++] = i;
        for (int i=0;i<n;i++) if (SA[i]>=k) tp[p++] = SA[i] - k;
        for (int i=0;i<m;i++) tax[i] = 0;
        for (int i=0;i<n;i++) tax[rk[tp[i]]]++;
        for (int i=1;i<m;i++) tax[i] += tax[i-1];
        for (int i=n-1;i>=0;i--) SA[--tax[rk[tp[i]]]] = tp[i];
        swap(tp, rk), p = 0, rk[SA[0]] = 0;
        for (int i=1;i<n;i++) rk[SA[i]] = cmp(tp, i, k) ? p : ++p;
        if (++p>=n) break;
        m = p;
    }
    int k = 0;
    for (int i=0;i<n;i++) {
        if (k) k--;
        int j = SA[rk[i]-1];
        while (a[i+k]==a[j+k]) k++;
        height[rk[i]] = k;
    }
}
bool check(int x) {
    int maxi , mini;
    maxi = mini = SA[1];
    for (int i=2;i<=n;i++) {
        if (height[i]>=x) {
            maxi = max(maxi, SA[i]);
            mini = min(mini, SA[i]);
        } else {
            if (mini+x<maxi) return true;
            maxi = mini = SA[i];
        } 
    }
    return false;
}
void clear() {}
void init() {
    clear();
    a[n] = 0;
    for (int i=0;i<n;i++) {
        scanf("%d", &a[i]);
        a[i-1] = a[i] - a[i-1] + 100;
    }
    a[n-1] = a[n] - a[n-1] + 100;
    n++, m = 300;
}
void solve() {
    build_SA();
    int l = 1, r = n, ans = 0;
    while (l<r) {
        int mid = (l+r)>>1;
        if (check(mid)) {
            ans = mid;
            l = mid + 1;
        } else r = mid;
    }
    if (ans<4) ans = 0; else ans++;
    printf("%d\n", ans);
}
int main() {
    #ifndef ONLINE_JUDGE
    freopen("1.in", "r", stdin);freopen("1.out", "w", stdout);
    #endif
    while (scanf("%d", &n)==1&&n) init(), solve();
    return 0;
}
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