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;

}

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