poj 1226 Substrings

最长公共子串（后缀数组解法）

仅正序

poj 2774 long long message

将字符串用不同的分隔符连接起来，求后缀数组和height数组即可，具体需要好好理解sa和height数组的含义。

正序逆序均可

poj 1226 Substrings

将字符串用分隔符与自己的逆序连接后，将所有串连接在一起，然后与正序相同的求法。

第二种情况的代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAX = 20010;

int s[MAX],t;
char ss[MAX];  
int n,k,size,pos[MAX],sp,vis[110];
int sa[MAX],rank[MAX],height[MAX];
int wa[MAX],wb[MAX],wv[MAX],ws[MAX];

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

void calheight(int *r, int n){  
   int i, j, k=0;  
   for(int i=1; i<=n; i++)  
       rank[sa[i]] = i;        
   for(int i=0; i<n; i++){
       if(k) k--;
       int j = sa[rank[i]-1];
       while(r[i+k] == r[j+k]) k++;
       height[rank[i]] = k;
    }   
}  

bool isgood(){
	for(int i=1; i<=size-3; i+=2){
		vis[i] += vis[i+1];
		if(!vis[i]) return false;
	}
	return true;
}

bool check(int mid){
	memset(vis, 0, sizeof(vis));
	int cnt = 0;
	for(int i=2; i<=k; i++){
		if(height[i] >=mid){
			if(!vis[pos[sa[i-1]]]){
				vis[pos[sa[i-1]]] = 1;
				cnt++;
			}
			if(!vis[pos[sa[i]]]){
				vis[pos[sa[i]]] = 1;
				cnt++;
			}
			if(cnt == n) return true;
		}
		else{
			cnt = 0;
			memset(vis, 0, sizeof(vis));
		}
	}	
	return false;
}

int main(){
	int cas;
	scanf("%d",&cas);
	while(cas--){
		t = 1000;
		scanf("%d",&n);
		k = 0, size = 0,sp = 240;
		for(int i=0; i<n; i++){
			scanf("%s",ss);
			int len = strlen(ss);
			t = min(len, t);
			for(int j=0; ss[j]; j++){
				pos[k] = i;
				s[k++] = ss[j];
			}
			pos[k] = sp;
			s[k++] = sp++;
			for(int j=len-1; j>=0; j--){
				pos[k] = i;
				s[k++] = ss[j];
			}
			pos[k] = sp;
			s[k++] = sp++;
		}
		s[k] = 0;
		fun(s, k+1, sp);
		calheight(s, k);
		int l = 0, r = t, ans = -1;
		while(l <= r){
			int mid = (l + r) >> 1;
			if(check(mid)){
				ans = mid;
				l = mid + 1;
			}
			else
				r = mid - 1;
		}
		printf("%d\n",ans);
	}
	return 0;
}