codeforces_374c

在一个二维表中搜特定序列的串。采用记忆化的搜索。

dp[x][y]表示x,y能达到的最大深度，如果每次搜的时候能搜到原来的点，那么就是无穷的。如果dp[x][y]的深度大于4，说明是存在特定串的。

AC代码：

#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

#define max(a,b) ((a) > (b)) ? (a) : (b)
const int MAX_NUMBER = 1003;
int n, m;
char next[] = "DIMA";
char maps[MAX_NUMBER][MAX_NUMBER];
int steps[][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int vis[MAX_NUMBER][MAX_NUMBER];
int dp[MAX_NUMBER][MAX_NUMBER];
int find, infi, max_ans;

void dfs(int cnt_x, int cnt_y, char cnt_char) {
    if (dp[cnt_x][cnt_y] != -1) {
        return ;
    }
    if (cnt_char == 'A') {
        dp[cnt_x][cnt_y] = 1;
        find = 1;
    }
    int next_char;
    for (int i = 0; i < 4; i++) {
        if (next[i] == cnt_char) {
            next_char = i;
            break;
        }
    }
    for (int i = 0; i < 4 && !infi; i++) {
        int next_x = cnt_x + steps[i][0];
        int next_y = cnt_y + steps[i][1];
        if (maps[next_x][next_y] == next[(next_char + 1) % 4]) {
            if (vis[next_x][next_y]) {
                infi = 1;
            }
            else {
                vis[next_x][next_y] = 1;
                dfs(next_x, next_y, next[(next_char + 1) % 4]);
                dp[cnt_x][cnt_y] = max(dp[cnt_x][cnt_y], dp[next_x][next_y] + 1);
                vis[next_x][next_y] = 0;
            }
        }
    }
}

int main() {

    scanf("%d%d", &n, &m);
    memset(maps, 0, sizeof(maps));
    for (int i = 1; i <= n; i++) {
        scanf("%s", maps[i] + 1);
    }
    memset(vis, 0, sizeof(vis));
    memset(dp, -1, sizeof(dp));
    find = infi = max_ans = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (maps[i][j] == 'D' && !vis[i][j]) {
                vis[i][j] = 1;
                dfs(i, j, 'D');
                if (max_ans < dp[i][j] / 4) {
                    max_ans = dp[i][j] / 4;
                }
                vis[i][j] = 0;
            }
        }
    }
    if (!find) {
        printf("Poor Dima!\n");
    }
    else {
        if (infi) {
            printf("Poor Inna!\n");
        }
        else {
            printf("%d\n", max_ans);
        }
    }
    return 0;
}