hdu2328(后缀数组 + 二分)

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2328

题意: 求 n 个串的字典序最小的最长公共子串

思路: 本题中单个字符串长度不超过 200, 可以暴力枚举一个字符串的所有前缀, 然后用kmp去匹配其他字符串.

我这里是用后缀数组写的. 类似 http://www.cnblogs.com/geloutingyu/p/7450580.html

不过本题是有 n 个字符串, 不能直接暴力判断. 不难想到这里可以直接二分答案长度, 不过 check 函数比较难想到, 具体看代码

代码:

#include <iostream>
#include <stdio.h>
#include <string.h>
#define rank Rank
using namespace std;

const int MAXN = 1e6 + 10;
int sol;
char str[MAXN];
int SA[MAXN], rank[MAXN], height[MAXN], sum[MAXN], tp[MAXN], vis[MAXN], tag[(int)(4e3 + 10)];

bool cmp(int *f, int x, int y, int w){
    return f[x] == f[y] && f[x + w] == f[y + w];
}

void DA(char *s, int n, int m){
    for(int i = 0; i < m; i++) sum[i] = 0;
    for(int i = 0; i < n; i++) sum[rank[i] = s[i]]++;
    for(int i = 1; i < m; i++) sum[i] += sum[i - 1];
    for(int i = n - 1; i >= 0; i--) SA[--sum[rank[i]]] = i;
    for(int len = 1; len <= n; len <<= 1){
        int p = 0;
        for(int i = n - len; i < n; i++) tp[p++] = i;
        for(int i = 0; i < n; i++){
            if(SA[i] >= len) tp[p++] = SA[i] - len;
        }
        for(int i = 0; i < m; i++) sum[i] = 0;
        for(int i = 0; i < n; i++) sum[rank[tp[i]]]++;
        for(int i = 1; i < m; i++) sum[i] += sum[i - 1];
        for(int i = n - 1; i >= 0; i--) SA[--sum[rank[tp[i]]]] = tp[i];
        swap(rank, tp);
        p = 1;
        rank[SA[0]] = 0;
        for(int i = 1; i < n; i++){
            rank[SA[i]] = cmp(tp, SA[i - 1], SA[i], len) ? p - 1 : p++;
        }
        if(p >= n) break;
        m = p;
    }
    int k = 0;
    n--;
    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 check(int mid, int n, int len){
    int cnt = 1;
    memset(tag, 0, sizeof(tag));
    tag[vis[SA[1]]] = 1;
    for(int i = 2; i <= len; i++){
        if(height[i] >= mid){
            if(!tag[vis[SA[i]]]){
                cnt++;
                tag[vis[SA[i]]] = 1;
                if(cnt >= n){
                    sol = SA[i];
                    return true;
                }
            }
        }else{
            cnt = 1;
            memset(tag, 0, sizeof(tag));
            tag[vis[SA[i]]] = 1;
        }
    }
    return false;
}

int main(void){
    int n;
    while(~scanf("%d", &n) && n){
        memset(vis, 0, sizeof(vis));
        scanf("%s", str);
        int len = strlen(str);
        int mx = len;
        for(int i = 1; i < n; i++){
            vis[len] = 1;
            str[len] = '0';
            scanf("%s", str + 1 + len);
            mx = max(mx, (int)(strlen(str) - len));
            len = strlen(str);
        }
        str[len] = 0;
        DA(str, len + 1, 'z' + 1);
        for(int i = 1; i <= len; i++) vis[i] += vis[i - 1];
        int l = 1, r = mx;
        while(l <= r){
            int mid = (l + r) >> 1;
            if(check(mid, n, len)) l = mid + 1;
            else r = mid - 1;
        }
        if(l - 1 <= 0){
            puts("IDENTITY LOST");
            continue;
        }
        for(int i = sol, j = 1; j <= l - 1; i++, j++){
            printf("%c", str[i]);
        }
        puts("");
    }
    return 0;
}