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  • HDU 5763 Another Meaning (kmp + dp)

    Another Meaning

    题目链接:

    http://acm.hdu.edu.cn/showproblem.php?pid=5763

    Description

    As is known to all, in many cases, a word has two meanings. Such as “hehe”, which not only means “hehe”, but also means “excuse me”.
    Today, ?? is chating with MeiZi online, MeiZi sends a sentence A to ??. ?? is so smart that he knows the word B in the sentence has two meanings. He wants to know how many kinds of meanings MeiZi can express.

    Input

    The first line of the input gives the number of test cases T; T test cases follow.
    Each test case contains two strings A and B, A means the sentence MeiZi sends to ??, B means the word B which has two menaings. string only contains lowercase letters.

    Limits
    T <= 30
    |A| <= 100000
    |B| <= |A|

    Output

    For each test case, output one line containing “Case #x: y” (without quotes) , where x is the test case number (starting from 1) and y is the number of the different meaning of this sentence may be. Since this number may be quite large, you should output the answer modulo 1000000007.

    Sample Input

    4
    hehehe
    hehe
    woquxizaolehehe
    woquxizaole
    hehehehe
    hehe
    owoadiuhzgneninougur
    iehiehieh

    Sample Output

    Case #1: 3
    Case #2: 2
    Case #3: 5
    Case #4: 1

    Hint

    In the first case, “ hehehe” can have 3 meaings: “he”, “he”, “hehehe”.
    In the third case, “hehehehe” can have 5 meaings: “hehe”, “hehe”, “hehe*”, “**”, “hehehehe”.

    Source

    2016 Multi-University Training Contest 4


    ##题意: 给出字符串A和B,单词B具有两种解释,求字符串A一共有多少种意义.
    ##题解: 考虑A中每个单词B,需要考虑它是否解释为第二种意义;很显然要用DP. 首先用kmp处理出A中的每个B,记录每个B的起点和终点(下面的代码记录的是以i为终点的B的起点,没有则为-1). dp[i]为处理到第i个字符时共有多少种意义: 若i是某个B的终点,则dp[i] = dp[i-1] + dp[起点-1]. (前者为当前B不解释为二义,后者为把B解释为二义). 若i不是终点,则dp[i] = dp[i-1];
    值得一提的是:由于背包dp需要用到初始值dp[0],而字符串又从0开始,导致初始值为dp[-1],所以下面的代码特别处理了这个问题. 实际上,应该在读入字符串时空出str[0]来避免这个问题.

    ##代码: ``` cpp #include #include #include #include #include #include #include #include #include #define LL int #define eps 1e-8 #define maxn 101000 #define mod 1000000007 #define inf 0x3f3f3f3f #define IN freopen("in.txt","r",stdin); using namespace std;

    char str[maxn],p[maxn];
    int f[maxn],cnt;
    int flag[maxn];
    void getf()
    {
    memset(f,0,sizeof(f));
    memset(flag, -1, sizeof(flag));
    int len=strlen(p);
    for(int i=1;i<len;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 ans;
    void kmp()
    {
    getf();
    int len1=strlen(str),len2=strlen(p);
    for(int i=0,j=0;i<len1;i++)
    {
    while(j&&str[i]!=p[j]) j=f[j];
    if(str[i]
    p[j]) j++;
    if(j==len2) {
    flag[i] = i-len2+1;
    ans++;
    }
    }
    }

    int dp[maxn];

    //由于字符串从0开始,这里需要用到dp[-1] = 1;
    int dps(int i) {
    if(i<0) return 1;
    return dp[i]%mod;
    }

    int main(int argc, char const *argv[])
    {
    //IN;

    int t; cin >> t; int ca = 1;
    while(t--)
    {
        ans = 0;
        scanf("%s", str); scanf("%s", p);
        kmp();
        fill(dp, dp+maxn, 1);
    
        int len = strlen(str);
        int len2 = strlen(p);
        for(int i=0; i<len; i++) {
            if(flag[i] == -1) dp[i] = dps(i-1)%mod;
            else {
                dp[i] = dps(i-1);
                dp[i] = (dp[i] + dps(flag[i]-1))%mod;
            }
        }
    
        printf("Case #%d: %d
    ", ca++, dp[len-1]%mod);
    }
    
    return 0;
    

    }

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  • 原文地址:https://www.cnblogs.com/Sunshine-tcf/p/5716585.html
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