UVA-705 Slash Maze

来源：互联网发布：ug nx软件编辑：程序博客网时间：2024/06/07 04:57

UVA-705 Slash Maze

题目大意：用/ \给出一个斜着的矩阵，求里面封闭环的个数和最大长度。
解题思路：一开始就懵逼系列，看同学题解有的思路，将给的图进行放大令 ‘\’ 变成
100
010
001
‘/’反之然后用dfs找被1包围0的个数除以 3即可

/*************************************************************************    > File Name: UVA-705.cpp    > Author: Robin    > Mail: 499549060@qq.com     > Created Time: 2016年08月08日 星期一 11时36分48秒 ************************************************************************/#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>#include<cstring>using namespace std;int d[4][2] = { {1,0}, {0,1}, {-1,0}, {0,-1}};int map[3000][3000];int m,n;int num;int flag;void dfs (int x, int y) {    map[x][y] = 1;    for (int i = 0; i < 4; i++)         if (x+d[i][0] >= 0 && x+d[i][0] < 3*n && y+d[i][1] >= 0 && y+d[i][1] < 3*m) {            if (map[x+d[i][0]][y+d[i][1]] == 0) {                num++;                dfs (x+d[i][0], y + d[i][1]);            }        }else flag = 0;}int main() {    int t = 1;    while (scanf("%d%d\n", &m, &n) && m+n) {        memset(map, 0, sizeof(map));        char c;        int s = 0;        for (int i = 0; i < n; i++)            for (int j = 0; j < m+1; j++) {                scanf("%c", &c);                if ( c == '\\') {                    map[3*i][3*j] = 1;                    map[3*i+1][3*j+1] = 1;                    map[3*i+2][3*j+2] = 1;                }                if (c == '/') {                    map[3*i][3*j+2] = 1;                    map[3*i+1][3*j+1] = 1;                    map[3*i+2][3*j] = 1;                }            }        int max = 0;        for (int i = 0; i < 3*n; i++)            for (int j = 0; j < 3*m; j++) {                flag = 1;                num = 1;                if (map[i][j] == 0) {dfs(i,j);                if ( flag) {                    s++;                    if ( num >= max) max = num;                }                }            }        if (s == 0) printf("Maze #%d:\nThere are no cycles.\n\n",t);        else printf("Maze #%d:\n%d Cycles; the longest has length %d.\n\n",t, s,max/3);        t++;    }}

0 0