HDU 5763 Another Meaning 多校赛 (DP + KMP)

Another Meaning

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1111    Accepted Submission(s): 518

Problem 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

4hehehehehewoquxizaolehehewoquxizaoleheheheheheheowoadiuhzgneninouguriehiehieh

 

Sample Output

Case #1: 3Case #2: 2Case #3: 5Case #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”, “he*he”, “hehe*”, “**”, “hehehehe”.
 

 

Author

FZU

 

Source

2016 Multi-University Training Contest 4

 

先KMP求出每一个子串在母串中的位置，然后用DP求解答案即可

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>using namespace std;typedef long long LL;const int MAXN = 100000 + 5;const int MAXM = 1000000 + 5;const int mod = 1000000007;char A[MAXN], B[MAXM];int T, nexts[MAXN];bool vis[MAXN];LL dp[MAXN];void Get_Next(int M) {    int i = 0,j = -1;    nexts[0] = -1;    while(i < M) {        if(j == -1 || B[i] == B[j]) {            i ++;            j ++;            if(B[i] == B[j]) {                nexts[i] = nexts[j];            } else {                nexts[i] = j;            }        } else {            j = nexts[j];        }    }}void Get_KMP(int N,int M) {    int i = 0,j = 0;    while(i < N && j < M) {        if(j == -1 || B[j] == A[i]) {            i ++;            j ++;        } else {            j = nexts[j];        }        if(j == M) {            vis[i - 1] = true;            j = nexts[j];        }    }}int main() {    scanf("%d", &T);    int cas = 1;    while(T --) {        scanf("%s%s", A, B);        memset(vis, false, sizeof(vis));        int t1 = strlen(A), t2 = strlen(B);        Get_Next(t2);        Get_KMP(t1, t2);        memset(dp, 0, sizeof(dp));        dp[0] = 1;        for(int i = 1; i <= t1; i ++) {            if(vis[i - 1]) dp[i] = dp[i - t2];            dp[i] = (dp[i] + dp[i - 1]) % mod;        }        printf("Case #%d: %lld\n",cas ++,  dp[t1]);    }    return 0;}