hdoj 5311 Hidden String 【KMP + 暴力】

Hidden String Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1434    Accepted Submission(s): 514 Problem Description Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string  s  of length  n . He wants to find three nonoverlapping substrings  s[l1..r1] ,  s[l2..r2] ,  s[l3..r3]  that: 1.  1≤l1≤r1<l2≤r2<l3≤r3≤n 2. The concatenation of  s[l1..r1] ,  s[l2..r2] ,  s[l3..r3]  is "anniversary".   Input There are multiple test cases. The first line of input contains an integer  T   (1≤T≤100) , indicating the number of test cases. For each test case: There's a line containing a string  s   (1≤|s|≤100)  consisting of lowercase English letters.   Output For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).   Sample Input 2 annivddfdersewwefary nniversarya   Sample Output YES NO

题意：给你一个字符串，让你从中找出s1[l1, r1]、s2[l2, r2]、s3[l3,r3]三段子串组成字符串anniversary，且三段子串满足l1 <= r1 <= l2 <= r2 <= l3 <= r3。问你能不能找到这样的三段子串。

思路：对字符串anniversary，每次枚举字符串中两个分割点i 和 j，分别构成三个相应的子串，查找在文本串中是否存在。

注意查找成功后，要重新构建文本串。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int f[10];
char ss[] = {"anniversary"};
char str[210], str1[210], str2[210];
void getfail(char *P)
{
    int l = strlen(P);
    f[0] = f[1] = 0;
    for(int i = 1; i < l; i++)
    {
        int j = f[i];
        while(j && P[i] != P[j])
            j = f[j];
        f[i+1] = P[i] == P[j] ? j + 1 : 0;
    }
}
int Find(char *T, char *P)//在字符串T中 查找字符串P
{
    int l1 = strlen(T);
    int l2 = strlen(P);
    int j = 0;
    for(int i = 0; i < l1; i++)
    {
        while(j && T[i] != P[j])
            j = f[j];
        if(T[i] == P[j])
            j++;
        if(j >= l2)
            return i+1;//返回最后匹配位置
    }
    return -1;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%s", str);
        bool flag = false;
        char s[10];
        int p, pos;
        for(int i = 1; i <= 9; i++)//第一分割点
        {
            if(flag)
                break;
            for(int j = i; j <= 9; j++)//第二分割点
            {
                //截取第一个字符串
                p = 0;
                for(int k = 0; k < i; k++)
                    s[p++] = ss[k];
                s[p] = '\0';
                getfail(s);//求失配函数
                if(Find(str, s) != -1)//查找
                {
                    pos = Find(str, s);
                    p = 0;
                    int len = strlen(str);
                    for(int k = pos; k < len; k++)//重构文本串
                        str1[p++] = str[k];
                    str1[p] = '\0';
                }
                else
                    continue;
                //截取第二个字符串
                p = 0;
                for(int k = i; k <= j; k++)
                    s[p++] = ss[k];
                s[p] = '\0';
                getfail(s);
                if(Find(str1, s) != -1)
                {
                    pos = Find(str1, s);
                    p = 0;
                    int len = strlen(str1);
                    for(int k = pos; k < len; k++)//重构文本串
                        str2[p++] = str1[k];
                    str2[p] = '\0';
                }
                else
                    continue;
                //截取第三个
                p = 0;
                for(int k = j+1; k < 11; k++)
                    s[p++] = ss[k];
                s[p] = '\0';
                getfail(s);
                if(Find(str2, s) != -1)
                {
                    flag = true;//第三个成功就 ok了
                    break;
                }
            }
        }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}