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Problem J: Island Buses

2024-11-20 11:37:02   203人阅读

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……

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#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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Problem J: Island Buses


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