POJ 3080 后缀数组/KMP

 

题目链接：http://poj.org/problem?id=3080

题意：给定n个DNA串，求最长公共子串。如果最长公共子串的长度小于3时输出no significant commonalities，否则输出该子串，如有多解请输出字典序最小的解

思路：是POJ 3405的弱化版。思路请参考

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<queue>
#include<vector>
#include<time.h>
#include<cmath>
#include<set>
using namespace std;
typedef long long int LL;
const int MAXN = 1000 + 10;
int wa[MAXN], wb[MAXN], wv[MAXN], WS[MAXN];
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++) WS[i] = 0;
    for (i = 0; i < n; i++) WS[x[i] = r[i]]++;
    for (i = 1; i < m; i++) WS[i] += WS[i - 1];
    for (i = n - 1; i >= 0; i--) sa[--WS[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++) WS[i] = 0;
        for (i = 0; i < n; i++) WS[wv[i]]++;
        for (i = 1; i < m; i++) WS[i] += WS[i - 1];
        for (i = n - 1; i >= 0; i--) sa[--WS[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;
}
int Rank[MAXN], height[MAXN], sa[MAXN];
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 r[MAXN], len, n, t, Index[MAXN], pos[60 + 5];
char str[60 + 5];
bool check(int x){
    set<int>se;
    for (int i = 1; i < len; i++){
        if (height[i] >= x){
            pos[x] = sa[i];
            se.insert(Index[sa[i]]); se.insert(Index[sa[i - 1]]);
        }
        else{
            if (se.size() == n){ return true; } se.clear();
        }
    }
    if (se.size() == n){ return true; }
    return false;
}
void solve(){
    if (n == 1){
        printf("%d\n", strlen(str));
        return;
    }
    int L = 1, R = 60, mid, ans = 0;
    while (R >= L){
        mid = (L + R) / 2;
        if (check(mid)){
            ans = mid;
            L = mid + 1;
        }
        else{
            R = mid - 1;
        }
    }
    if (ans <3){
        printf("no significant commonalities\n");
    }
    else{
        for (int i = pos[ans], j = 0; j < ans; j++, i++){
            printf("%c", (r[i] - n - 1) + 'A');
        }
        printf("\n");
    }
}
int main(){
    //#ifdef kirito
    //    freopen("in.txt", "r", stdin);
    //    freopen("out.txt", "w", stdout);
    //#endif
    //    int start = clock();
    scanf("%d", &t);
    while (t--){
        scanf("%d", &n); len = 0;
        for (int i = 1, val = 0; i <= n; i++){
            scanf("%s", &str);
            for (int j = 0; j < strlen(str); j++){
                Index[len] = i;
                r[len++] = (str[j] - 'A' + n + 1);
            }
            Index[len] = i;
            r[len++] = val++;
        }
        da(r, sa, len, 100);
        calheight(r, sa, len - 1);
        solve();
    }
    //#ifdef LOCAL_TIME
    //    cout << "[Finished in " << clock() - start << " ms]" << endl;
    //#endif
    return 0;
}