hdu4436(后缀自动机)

最新推荐文章于 2019-07-23 00:04:20 发布

原创最新推荐文章于 2019-07-23 00:04:20 发布 · 593 阅读

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这篇博客探讨了如何使用后缀自动机来解决计算多个字符串组成的不同数的个数，并对这些数进行去重求和的问题。通过在字符串间插入特殊字符以确保在构建自动机时不会出现特定路径。

题解：求多个字符串能够构成多少种数然后把这些数去重后加起来，后缀自动机有个性质就是路径数就等于不同子串个数，然后怎么把多个字符串拼接呢，两个字符串中间加入一个你不会用的字符，因为这样可以保证我到时候路径不会走这条

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<cstdlib>
#include<ctime>
#include<stack>
#include<bitset>
using namespace std;
#define mes(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i = a; i <= b; i++)
#define dec(i,a,b) for(int i = b; i >= a; i--)
#define fi first
#define se second
#define ls rt<<1
#define rs rt<<1|1
#define lson ls,L,mid
#define rson rs,mid+1,R
#define lowbit(x) x&(-x)
typedef double db;
typedef long long int ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const ll inf = 0x3f3f3f3f;
const int mx = 2e5+3e4;
const int mod = 1e9+7;
const int x_move[] = {1,-1,0,0,1,1,-1,-1};
const int y_move[] = {0,0,1,-1,1,-1,1,-1};
int n,m;
struct sam{
    int ch[mx][20];
    int len[mx];
    int id[mx];
    int f[mx];
    int cnt[mx];
    bool vis[mx];
    int sum[mx];
    int c[mx];
    int sz,rt,last;
    int newnode(int val){
        ++sz;
        len[sz] = val;
        mes(ch[sz],0);
        return sz;
    }
    void init(){
        sz = 0;
        last = rt = newnode(0);
        f[rt] = 0;
    }
    void insert(int c){
        int p = last,np = newnode(len[last]+1);
        vis[np] = (c==10);
        while(p&&!ch[p][c]) ch[p][c] = np,p = f[p];
        if(!p)
            f[np] = rt;
        else{
            int q = ch[p][c];
            if(len[q]==len[p]+1) f[np] = q;
            else{
                    int nq = newnode(len[p]+1);
                    memcpy(ch[nq],ch[q],sizeof(ch[q]));
                    vis[nq] = vis[q];
                    f[nq] = f[q];
                    f[q] = f[np] = nq;
                    while(p&&ch[p][c]==q) ch[p][c] = nq,p = f[p];
            }
        }
        last = np;
    }
    void rsort(){
        for(int i = 0; i <= sz; i++) sum[i] = cnt[i] = c[i] = 0;
        for(int i = 1; i <= sz; i++) c[len[i]]++;
        for(int i = 1; i <= sz; i++) c[i] += c[i-1];
        for(int i = sz; i >= 1; i--) id[c[len[i]]--] = i;
    }
    void solve(){
        int ans = 0;
        cnt[1] = 1;
        for(int i = 1; i <= sz; i++){
            int j = id[i];
            ans += sum[j];
            ans %= 2012;
            for(int d = 0; d < 10; d++)if(ch[j][d]&&(d!=0||len[j]!=0)){
                int k = ch[j][d];
                sum[k] += sum[j]*10+cnt[j]*d;    
                cnt[k] += cnt[j];
                sum[k]%=2012;
                cnt[k]%=2012;
            }
        }
        printf("%d\n",ans);
    }
}word;
char str[mx];
int main(){
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    int t,q,ca = 1;
    while(~scanf("%d",&n)){
        word.init();
        for(int i = 1; i <= n; i++){
            scanf("%s",str+1);
            int j;
            for(j = 1; str[j]=='0'; j++);
            for( ; str[j]; j++)
                word.insert(str[j]-'0');
            word.insert(10);
        }
        word.rsort();
        word.solve();
    }
    return 0;
}