HDU 5024 Wang Xifeng's Little Plot (DP)

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;
}