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;

}