HDU 5763 Another Meaning KMP+DP
2016-08-06 10:56
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题目链接:
http://acm.hdu.edu.cn/showproblem.php?pid=5763Another Meaning
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)问题描述
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.
输入
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|
输出
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.
样例
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
题解
dp
用kmp处理出匹配成功的结尾的字母位置,然后就是考虑每个位置选和不选的两种情况,转移下就可以了。
代码
#include<map> #include<queue> #include<vector> #include<cstdio> #include<cstring> #include<string> #include<iostream> #include<algorithm> #define X first #define Y second #define mkp make_pair #define lson (o<<1) #define rson ((o<<1)|1) #define M (l+(r-l)/2) #define bug(a) cout<<#a<<" = "<<a<<endl using namespace std; typedef __int64 LL; const int maxn=1e5+10; const int INF=0x3f3f3f3f; const LL INFL=0x3f3f3f3f3f3f3f3fLL; const int mod=1e9+7; char s1[maxn],s2[maxn]; int vis[maxn]; vector<int> pos; LL dp[maxn]; int f[maxn]; void getFail(char *P){ int m=strlen(P); f[0]=0; f[1]=0; for(int i=1;i<m;i++){ int j=f[i]; while(j&&P[i]!=P[j]) j=f[j]; f[i+1]=P[i]==P[j]?j+1:0; } } void find(char* T,char *P){ int n=strlen(T),m=strlen(P); getFail(P); int j=0; for(int i=0;i<n;i++){ while(j&&P[j]!=T[i]) j=f[j]; if(P[j]==T[i]) j++; if(j==m){ pos.push_back(i+1); vis[i+1]=1; } } } void init(){ pos.clear(); memset(vis,0,sizeof(vis)); } int main() { int tc,kase=0; scanf("%d",&tc); while(tc--){ init(); scanf("%s%s",s1,s2); find(s1,s2); memset(dp,0,sizeof(dp)); dp[0]=1; int n=strlen(s1),m=strlen(s2); for(int i=1;i<=n;i++){ //第i位不选 dp[i]=dp[i-1]; //第i位选 if(vis[i]) dp[i]+=dp[i-m]; dp[i]%=mod; } printf("Case #%d: %I64d\n",++kase,dp ); } return 0; }
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