poj 3050 dfs + set的妙用

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题意:

给一个5x5的矩阵，求由多少个由连续6个元素组成的不一样的字符的个数。

解析：

dfs + set去重搞定。

代码：

#include <iostream>#include <cstdio>#include <set>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <stack>#include <vector>#include <queue>#include <map>#include <climits>#include <cassert>#define LL long longusing namespace std;const int maxn = 300 + 10;const int inf = 0x3f3f3f3f;const double eps = 1e-8;const double pi = acos(-1.0);const double ee = exp(1.0);int dir[][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};set<int> ans;int g[10][10];void dfs(int x, int y, int sum, int dep){    if (dep == 6)    {        ans.insert(sum);        return;    }    for (int i = 0; i < 4; i++)    {        int nx = x + dir[i][0];        int ny = y + dir[i][1];        if (1 <= nx && nx <= 5 && 1 <= ny && ny <= 5)        {            dfs(nx, ny, sum * 10 + g[x][y], dep + 1);        }    }    return;}int main(){#ifdef LOCAL    freopen("in.txt", "r", stdin);#endif // LOCAL    ans.clear();    for (int i = 1; i <= 5; i++)    {        for (int j = 1; j <= 5; j++)        {            scanf("%d", &g[i][j]);        }    }    for (int i = 1; i <= 5; i++)    {        for (int j = 1; j <= 5; j++)        {            dfs(i, j, 0, 0);        }    }    printf("%d\n", ans.size());    return 0;}

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