Problem J: Island Buses

主要题意是：大海之间有岛，有的岛之间有桥，问你岛的个数，桥的个数，以及没有桥联通岛的个数，其中最后一次输入的没有回车，不注意的话最后一次会被吞，第二，桥的两端的标记是“X”（X也代表陆地），“X”的四周都可以有“B”形成的桥，一开始没写好，后来根据“X”标记所有的桥只能走一次然后标记……总之，虽然是水题，写出来还是蛮开心的……

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <map>
#include <cmath>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <cctype>

const double Pi = atan(1) * 4;
using namespace std;
char str[100][100];
bool visit1[100][100];
bool visit2[100][100];
int cnt ;
int len;
int bridge;
int dr[] = {1,-1,0,0};
int dc[] = {0,0,-1,1};
void dfs1(int r,int c){
    visit1[r][c] = 1;
    for(int i = 0;i < 4;i++){
        int xx = r + dr[i];
        int yy = c + dc[i];
        if(xx >= 0 && yy >= 0 && xx < cnt && yy < len){
            if(!visit1[xx][yy] && (str[xx][yy] == ‘#‘ || str[xx][yy] == ‘X‘ )){
                dfs1(xx,yy);
            }
        }
    }
}
void dfs2(int r,int c){
    visit2[r][c] = 1;
    for(int i = 0;i < 4;i++){
        int xx = r + dr[i];
        int yy = c + dc[i];
        if(xx >= 0 && yy >= 0 && xx < cnt && yy < len){
            if(!visit2[xx][yy] && (str[xx][yy] == ‘#‘ || str[xx][yy] == ‘X‘)){
                dfs2(xx,yy);
            }
            else if(str[xx][yy] == ‘B‘ && str[r][c] == ‘X‘ && !visit2[xx][yy]){
                int j = 0;
                visit2[xx][yy] = 1;
                bridge++;
                while(1){
                    j++;
                    int tt1 = xx + j * dr[i];
                    int tt2 = yy + j *  dc[i];
                    if(tt1 < 0 || tt2 < 0 || tt1 >= cnt || tt2 >= len)
                        break;
                    visit2[tt1][tt2] = 1;
                    if(str[tt1][tt2] == ‘X‘){
                        dfs2(tt1,tt2);
                        break;
                    }
                }
            }
        }
    }
}
int main()
{
    //freopen("input.in","r",stdin);
    //freopen("output.in","w",stdout);
    cnt = 0;
    int cas = 1;
    memset(str,0,sizeof(str));
    while(fgets(str[0],sizeof(str[0]),stdin) != NULL){
        if(cas != 1)
            cout << endl;
        len = strlen(str[0]) - 1;
        while((fgets(str[++cnt],sizeof(str[0]),stdin) )!= NULL){
            if(str[cnt][0] == 10){
                break;
            }
        }
        bridge = 0;
        int bus = 0;
        int island = 0;
        memset(visit1,0,sizeof(visit1));
        memset(visit2,0,sizeof(visit2));
        for(int i = 0;i <= cnt;i++){
            for(int j = 0;j < len;j++){
                if( (str[i][j] == ‘#‘ || str[i][j] == ‘X‘) && !visit1[i][j]){
                    island++;
                    dfs1(i,j);
                }
                if( (str[i][j] == ‘#‘ || str[i][j] == ‘X‘)&& !visit2[i][j]){
                    bus++;
                    dfs2(i,j);
                }
            }
        }
        cout << "Map " << cas++ << endl;
        cout << "islands: " << island << endl;
        cout << "bridges: " << bridge << endl;
        cout << "buses needed: " << bus << endl;
        cnt = 0;
        memset(str,0,sizeof(str));
    }
    return 0;
}