UVa 11127 - Triple-Free Binary Strings (位运算 + DFS)

题意

计算给定的字符串中有几个符合条件的字符串K,满足K的所有子串不包含三个相同的字符串连起来的字符串。

思路

一开始枚举1024种状态,放到set里,跑得慢不说还WA了。

还是看了帆神的题解,直接用位运算来判断。

其实我奇怪的是为什么30个*不会超时。

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <iomanip>
#include <cmath>
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const double eps = 1e-8;
const int MAXN = 10000 + 10;
const int MOD = 1000007;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
 
int len;
char str[50];
 
bool Judge(int state, int len)
{
    int pivot = (1 << len) - 1;
    int a = (state & pivot);
    state >>= len;
    int b = (state & pivot);
    state >>= len;
    if (a == b && b == state) return true;
    return false;
}
 
int DFS(int state, int pos)
{
    int tmpS = state;
    for (int i = 0; i <= pos - 3; i++)
    {
        if ((pos - i) % 3 == 0 && Judge(tmpS, (pos - i) / 3)) return 0;
        tmpS >>= 1;
    }
    if (pos == len) return 1;
    int ans = 0;
    if (str[pos] == '0') ans += DFS(state, pos + 1);
    else if (str[pos] == '1') ans += DFS(state | (1 << pos), pos + 1);
    else
    {
        ans += DFS(state, pos + 1);
        ans += DFS(state | (1 << pos), pos + 1);
    }
    return ans;
}
 
int main()
{
   // ROP;
    int n, i, j, cases = 0;
    while (scanf("%d", &n), n)
    {
        scanf("%s", str);
        len = strlen(str);
        printf("Case %d: %d\n", ++cases, DFS(0, 0));
    }
    return 0;
}

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