HDU 5024 Wang Xifeng's Little Plot (DP)

时间：2022-02-27 01:18:47

题意：给定一个n*m的矩阵，#表示不能走，.表示能走，让你求出最长的一条路，并且最多拐弯一次且为90度。

析：DP，dp[i][j][k][d] 表示当前在(i, j)位置，第 k 个方向，转了 d 次变的最多次数，然后用记忆化搜索就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 100 + 5;

const int mod = 1e9 + 7;

const int dr[] = {0, 1, 0, -1, -1, 1, 1, -1};

const int dc[] = {1, 0, -1, 0, 1, 1, -1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

int dp[maxn][maxn][10][2];

char s[maxn][maxn];

int dfs(int r, int c, int d, int num){

    int &ans = dp[r][c][d][num];

    if(ans >= 0)  return ans;

    ans = 1;

    int x = r + dr[d];

    int y = c + dc[d];

    if(is_in(x, y) && s[x][y] == '.')  ans = Max(ans, dfs(x, y, d, num) + 1);

    if(d < 4 && !num){

        x = r + dr[(d+1)%4];

        y = c + dc[(d+1)%4];

        if(is_in(x, y) && s[x][y] == '.')  ans = Max(ans, dfs(x, y, (d+1)%4, 1) + 1);

        x = r + dr[(d+3)%4];

        y = c + dc[(d+3)%4];

        if(is_in(x, y) && s[x][y] == '.')  ans = Max(ans, dfs(x, y, (d+3)%4, 1) + 1);

    }

    else if(!num){

        int t = (d + 1) % 8;

        if(t < 4)  t += 4;

        x = r + dr[t];

        y = c + dc[t];

        if(is_in(x, y) && s[x][y] == '.')  ans = Max(ans, dfs(x, y, t, 1) + 1);

        t = (d + 3) % 8;

        if(t < 4)  t += 4;

        x = r + dr[t];

        y = c + dc[t];

        if(is_in(x, y) && s[x][y] == '.')  ans = Max(ans, dfs(x, y, t, 1) + 1);

    }

    return ans;

}

int main(){

    while(scanf("%d", &n) == 1 && n){

        for(int i = 0; i < n; ++i)  scanf("%s", s+i);

        memset(dp, -1, sizeof dp);

        m = n;

        int ans = 0;

        for(int i = 0; i < n; ++i)

            for(int j = 0; j < n; ++j)

                if(s[i][j] == '.') for(int k = 0; k < 8; ++k)

                    ans = Max(ans, dfs(i, j, k, 0));

        printf("%d\n", ans);

    }

    return 0;

}

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