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?

InputEach 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.

OutputFor 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;}