hdu 5311 Hidden String dp o(n)算法 深搜

Hidden String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 857    Accepted Submission(s): 322

Problem Description

Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string s of length n . He wants to find three nonoverlapping substrings s[l1..r1] , s[l2..r2] , s[l3..r3] that:

1. 1≤l1≤r1<l2≤r2<l3≤r3≤n

2. The concatenation of s[l1..r1] , s[l2..r2] , s[l3..r3] is "anniversary".

 

Input

There are multiple test cases. The first line of input contains an integer T (1≤T≤100) , indicating the number of test cases. For each test case:

There's a line containing a string s (1≤|s|≤100) consisting of lowercase English letters.

 

Output

For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).

 

Sample Input

   
   
   
   

    
    
    
    2
annivddfdersewwefary
nniversarya
   
   
   
   

 

题意，就是给一段字符串，要求找出目标串，且最多只能三个片段串。

由于s很小，所以用深搜，暴力乱搞就好了。但这里用dp也能解决。复杂度是o(n)

定义pre[i][j] 表示字符串第i个位置，从目标串往后能匹配pre[i][j] 个，也就是说s(i,i+pre[i][j] ) == g(j,j+pre[i][j] ) g表示目标串。

则pre[i][j] 转状转移很简单就是pre[i+1][j+1]转移来就可以了。

定义dp[i][j]表示，字符串第i-1个位置前已经匹配到了目标串g的第j-1个位置且已用的分段数就是dp[i][j]。则状态转移方程就是

dp[i][j]可以从dp[i-1][j]得来，也可以转移到dp[i+pre[i][j]][j+pre[i][j]]，取分段数的最小值。

答案只要dp[s][11]大于1，说明存在。

由于本题目标串g的长度为定值，所以复杂度为o(s * 11)线性算法，如果，改变一下，g不是字长，而是变长的，那么复杂度则为o(s* g)了，与分段数是无关的。

#define N 205
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,glen,len,dp[N][12],pre[N][11];
char str[N];
char goal[20] = "anniversary";
void Update(int & a,int b){
    if(a == 0) a = b;
    else a = min(a,b);
}
int main()
{
    glen = strlen(goal);
    while(S(n)!=EOF)
    {
        while(n--){
            SS(str);len = strlen(str);
            FI(len){
                FJ(11){
                    if(str[i] == goal[j]) pre[i][j] = 1;
                    else pre[i][j] = 0;
                }
            }
            for(int i = len - 2;i>=0;i--){
                FJ(10){
                    if(pre[i][j] && pre[i+1][j+1]) pre[i][j] = pre[i+1][j+1] + 1;
                }
            }
            fill(dp,0);
            for(int i = 0;i<len;i++){
                FJ(11){
                    if(i && dp[i-1][j])   Update(dp[i][j],dp[i-1][j]);
                    if((dp[i][j] || !j )&&dp[i][j] < 3 && pre[i][j]){
                     Update(dp[i+pre[i][j]][j+pre[i][j]],dp[i][j] + 1);
                    }
                }
            }
            bool flag = true;
            for(int i = 0;i<=len && flag;i++){
                if(dp[i][11]) flag = false;
            }
            if(!flag)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    return 0;
}

再给个用深搜暴力解决的，更简单，更好理解，也好写，在比赛的时候，抢时间，也能有用处，但复杂度就很高了。

#define N 205
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,glen,len;
char str[N];
char goal[20] = "anniversary";
bool DFS(int si,int gi,int num){
    if(num <=3 && gi >= glen) return true;
    if(num > 3) return false;
    for(int i = si;i<len;i++){
        int ti = i,tgi = gi;
        while(ti<len && tgi < glen && str[ti] == goal[tgi]) tgi++,ti++;
        if(ti > i && DFS(ti,tgi,num + 1)) return true;
    }
    return false;
}
int main()
{
    glen = strlen(goal);
    while(S(n)!=EOF)
    {
        while(n--){
            SS(str);len = strlen(str);
            if(DFS(0,0,0))
            printf("YES\n");
            else
            printf("NO\n");
        }
    }
    return 0;
}

Sample Output

   
   
   
   

    
    
    
    YES
NO
   
   
   
   

 

Source

BestCoder 1st Anniversary ($)

 

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