HDU 4436 str2int （后缀自动机SAM,多串建立）

str2int

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 841    Accepted Submission(s): 297

Problem Description

In this problem, you are given several strings that contain only digits from '0' to '9', inclusive.
An example is shown below.
101
123
The set S of strings is consists of the N strings given in the input file, and all the possible substrings of each one of them.
It's boring to manipulate strings, so you decide to convert strings in S into integers.
You can convert a string that contains only digits into a decimal integer, for example, you can convert "101" into 101, "01" into 1, et al.
If an integer occurs multiple times, you only keep one of them. 
For example, in the example shown above, all the integers are 1, 10, 101, 2, 3, 12, 23, 123.
Your task is to calculate the remainder of the sum of all the integers you get divided by 2012.

 

 

Input

There are no more than 20 test cases.
The test case starts by a line contains an positive integer N.
Next N lines each contains a string consists of one or more digits.
It's guaranteed that 1≤N≤10000 and the sum of the length of all the strings ≤100000.
The input is terminated by EOF.

 

 

Output

An integer between 0 and 2011, inclusive, for each test case.

 

 

Sample Input

5 101 123 09 000 1234567890

 

 

Sample Output

202

 

 

Source

2012 Asia Tianjin Regional Contest

 

 

Recommend

zhoujiaqi2010

 

 

 

使用SAM进行多串建立。

 

然后拓扑排序，

之后累加计数就可以了。

注意前导0的要去掉。

 

/* ***********************************************
Author        :kuangbin
Created Time  :2013-10-11 17:50:31
File Name     :E:\2013ACM\专题强化训练\区域赛\2012天津\F.cpp
************************************************ */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MOD = 2012;
const int CHAR = 10;
const int MAXN = 100010;
struct SAM_Node
{
    SAM_Node *fa, *next[CHAR];
    int len;
    int id,pos;
    int cnt;
    int sum;
    SAM_Node(){}
    SAM_Node(int _len)
    {
        fa = 0;
        len = _len;
        memset(next,0,sizeof(next));
        cnt = sum = 0;
    }
};
SAM_Node SAM_node[MAXN*2], *SAM_root, *SAM_last;
int SAM_size;
SAM_Node *newSAM_Node(int len)
{
    SAM_node[SAM_size] = SAM_Node(len);
    SAM_node[SAM_size].id = SAM_size;
    return &SAM_node[SAM_size++];
}
SAM_Node *newSAM_Node(SAM_Node *p)
{
    SAM_node[SAM_size] = *p;
    SAM_node[SAM_size].id = SAM_size;
    SAM_node[SAM_size].cnt = SAM_node[SAM_size].sum = 0;
    return &SAM_node[SAM_size++];
}

void SAM_init()
{
    SAM_size = 0;
    SAM_root = SAM_last = newSAM_Node(0);
    SAM_node[0].pos = 0;
}
void SAM_add(int x,int len)
{
    SAM_Node *p = SAM_last, *np = newSAM_Node(p->len+1);
    np->pos = len;
    SAM_last = np;
    for(;p && !p->next[x];p = p->fa)
        p->next[x] = np;
    if(!p)
    {
        np->fa = SAM_root;
        return;
    }
    SAM_Node *q = p->next[x];
    if(q->len == p->len + 1)
    {
        np->fa = q;
        return;
    }
    SAM_Node *nq = newSAM_Node(q);
    nq->len = p->len + 1;
    q->fa = nq;
    np->fa = nq;
    for(;p && p->next[x] == q;p = p->fa)
        p->next[x] = nq;
}
//多串的建立,注意SAM_init()的调用
void SAM_build(char *s)
{
    int len = strlen(s);
    SAM_last = SAM_root;
    for(int i = 0;i < len;i++)
    {
        if( !SAM_last->next[s[i] - '0'] || !(SAM_last->next[s[i] - '0']->len == i+1) )
            SAM_add(s[i] - '0',i+1);
        else SAM_last = SAM_last->next[s[i] - '0'];
    }
}

char str[MAXN];
int topocnt[MAXN];
SAM_Node *topsam[MAXN*2];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int n;
    while(scanf("%d",&n) == 1)
    {
        SAM_init();
        for(int i = 0;i < n;i++)
        {
            scanf("%s",str);
            SAM_build(str);
        }
        //continue;
        memset(topocnt,0,sizeof(topocnt));
        for(int i = 0;i < SAM_size;i++)
            topocnt[SAM_node[i].len]++;
        for(int i = 1;i < MAXN;i++)
            topocnt[i] += topocnt[i-1];
        for(int i = 0;i < SAM_size;i++)
            topsam[--topocnt[SAM_node[i].len]] = &SAM_node[i];
        int ans = 0;
        SAM_root->cnt = 1;
        for(int i = 0;i < SAM_size;i++)
        {
            SAM_Node *tmp = topsam[i];
            for(int j = 0;j < 10;j++)
            {
                if(i == 0 && j == 0)continue;
                if(tmp->next[j])
                {
                    SAM_Node *q = tmp->next[j];
                    q->cnt = (q->cnt + tmp->cnt)%MOD;
                    q->sum = (q->sum + tmp->sum*10+tmp->cnt*j)%MOD;
                }
            }
            ans = (ans + tmp->sum)%MOD;
        }
        printf("%d\n",ans);

    }
    return 0;
}