[String processing + motion gauge] word Division

[String processing + motion gauge] word Division

Time Limit: 1000 ms
Memory limit: 2560kb

There is a long string consisting of lowercase letters. To facilitate the analysis of this string, you need to divide it into several parts, each of which is called a word. To reduce the analysis volume, we hope that the fewer words we divide, the better. You are here to complete this division.

Input Format
The first line is a string. (The length of a string cannot exceed 100)
The second row contains an integer N, indicating the number of words. (N <= 100)
3rd ~ N + 2 rows. Each row lists one word.

Output Format
An integer that represents the minimum number of words that a string can be divided.

Sample Input
Realityour
5
Real
Reality
It
Your
Our

Sample output
2
(The original string can be split into real + It + Your or reality + our. Since reality + our is only two parts, the optimal solution is 2. Note that, each word in the word list can be used multiple times or not)

Status: F [I] indicates the minimum number of splits to the end of F [I ].

State transition equation: If (F [I] = 0) f [I] = f [m] + 1; else f [I] = min (F [I], f [m] + 1 );

The data for this question is weak.

It is in the same status as a hungry ox.

# include<stdio.h># include<cstring># include<vector># include<iostream># include<algorithm>using namespace std;const int maxn=1000;vector<int>q[300];char st[maxn],dp[maxn][maxn];int f[maxn];int check(int a,int b,int c){    if(c-dp[q[a][b]][0]+1<=0)return 0;    for(int i=1,j=c-dp[q[a][b]][0]+1;i<=dp[q[a][b]][0]&&j>=1;i++,j++)    if(dp[q[a][b]][i]!=st[j])return 0;    return 1;}int main(){    scanf("%s\n",st+1);     st[0]=strlen(st+1);    int n;scanf("%d",&n);    for(int i=1;i<=n;i++){        scanf("%s",dp[i]+1);        dp[i][0]=strlen(dp[i]+1);        q[dp[i][dp[i][0]]].push_back(i);    }    for(int i=1;i<=st[0];i++)    for(int j=0;j<q[st[i]].size();j++)    if(check(int(st[i]),j,i)){    int m=i-dp[q[st[i]][j]][0];    if(f[i]==0)f[i]=f[m]+1;    else f[i]=min(f[i],f[m]+1);    }    printf("%d",f[st[0]]);    return 0;}

 

[String processing + motion gauge] word Division