sdnu oj 1136 Balloons bfs

注意每个答案后面要在空一行哦
经典bfs
题意是这样的（不知道图片为什么这么大…）

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <math.h>
#include <algorithm>
#include <queue>
#define mod 1000000007
using namespace std;

typedef long long ll;
const int N = 106;
const int inf = 0x3f3f3f3f;
const double eps = 1e-5;

struct node
{
    int x, y;
}one[N*N];

bool vis[N][N];
char a[N][N];
int der1[4][2] = {{-1,0},{1,0},{0,1},{0,-1}};
int der2[8][2] = {{-1,0},{1,0},{0,1},{0,-1},{-1,-1},{-1,1},{1,1},{1,-1}};
int cnt, ans1, ans2, n;

void bfs1()
{
    int i, j;
    queue<node> q;
    for(i = 1; i <= cnt; ++i)
    {
        if(!vis[one[i].x][one[i].y])         //从未遍历过的 '1' 找起， 可以作为一个连通块
        {
            ++ans1;
            q.push(one[i]);
            vis[one[i].x][one[i].y] = 1;
            while(!q.empty())
            {
                node now = q.front();
                q.pop();
                for(j = 0; j < 4; ++j)
                {
                    int xx = now.x + der1[j][0];
                    int yy = now.y + der1[j][1];
                    if(xx >= 0 && xx < n && yy >= 0 && yy < n)
                        if(!vis[xx][yy] && a[xx][yy] == '1')
                        {
                            vis[xx][yy] = 1;
                            node tp;
                            tp.x = xx;
                            tp.y = yy;
                            q.push(tp);
                        }
                }
            }
        }
    }
    printf("%d", ans1);
}

void bfs2()
{
    int i, j;
    queue<node> q;
    for(i = 1; i <= cnt; ++i)
    {
        if(!vis[one[i].x][one[i].y])         
        {
            ++ans2;
            q.push(one[i]);
            vis[one[i].x][one[i].y] = 1;
            while(!q.empty())
            {
                node now = q.front();
                q.pop();
                for(j = 0; j < 8; ++j)
                {
                    int xx = now.x + der2[j][0];
                    int yy = now.y + der2[j][1];
                    if(xx >= 0 && xx < n && yy >= 0 && yy < n)
                        if(!vis[xx][yy] && a[xx][yy] == '1')
                        {
                            vis[xx][yy] = 1;
                            node tp;
                            tp.x = xx;
                            tp.y = yy;
                            q.push(tp);
                        }
                }
            }
        }
    }
    printf(" %d\n\n", ans2);
}
int main()
{
    int i, j;
    int num = 0;
    while(~scanf("%d", &n) && n)
    {
        getchar();      //“吃掉”换行回车
        ++num;
        ans1 = ans2 = cnt = 0;
        memset(vis, 0, sizeof vis);
        memset(one, 0, sizeof one);
        memset(a, '\0', sizeof a);     
        for(i = 0; i < n; ++i)
        {
            for(j = 0; j < n; ++j)
            {
                scanf("%c", &a[i][j]);
                if(a[i][j] == '1')          //记录'1'出现的位置
                {
                    one[++cnt].x = i;
                    one[cnt].y = j;
                }
            }
            getchar();         //“吃掉”换行回车
        }
        printf("Case %d: ", num);
        bfs1();
        memset(vis, 0, sizeof vis);
        bfs2();
    }
    return 0;
}