uva 11127 - Triple-Free Binary Strings

Problem J
Triple-Free Binary Strings 
Input: Standard Input

Output: Standard Output

 

A binary string consists of ones and zeros. Given a binary string T, if  there is no binary string S such that SSS (concatenate three copies of S together) is a substring of T, we say T is triple-free.

A pattern consists of ones, zeros and asterisks, where an asterisk(*) can be replaced by either one or zero. For example, the pattern 0**1 contains strings 0001, 0011, 0101, 0111, but not 1001 or 0000.

Given a pattern P, how many triple-free binary strings does it contain?

Input
Each line of the input represents a test case, which contains the length of pattern, n(0<n<31), and the pattern P. There can be maximum 35 test cases. 

The input terminates when n=0.

Output
For each test case, print the case number and the answer, shown below. 

 

Sample Input                       Output for Sample Input

4 0**1 5 ***** 10 **01**01** 0

Case 1: 2 Case 2: 16 Case 3: 9

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Problemsetter: Rujia Liu
Special Thanks: Shahriar Manzoor

 枚举出所有的字符串，再判断肯定要超时。所以需要剪枝，就是末尾每多考虑一个字符，就判断因这个字符是否会引起不合法，不合法则剪枝。

#include <cstdio>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <iostream>
#include <stack>
#include <set>
#include <cstring>
#include <stdlib.h>
#include <cmath>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 100+5;
const int INF = 1000000000;

char s[maxn];
int n;

bool judge(int d){
    for(int i = 1;i*3 <= (d+1);i++){
        int tag = 1;
        for(int j = d;j >= d-i+1;j--){
            if(s[j] != s[j-i] || s[j] != s[j-2*i]){
                tag = 0;
                break;
            }
        }
        if(tag == 1) return false;
    }
    return true;
}

int dfs(int d){
    if(d == n){
        return 1;
    }
    int ret = 0;
    if(s[d] == '*'){
        s[d] = '0';
        if(judge(d))
            ret += dfs(d+1);
        s[d] = '1';
        if(judge(d))
            ret += dfs(d+1);
        s[d] = '*';
    }
    else{
        if(judge(d))
            ret += dfs(d+1);
    }
    return ret;
}

int main(){
    int kase = 0;
    while(scanf("%d", &n)){
        kase++;
        if(n == 0) break;
        scanf("%s", s);
        printf("Case %d: %d\n", kase, dfs(0));
    }
    return 0;
}