【题目链接】:http://hihocoder.com/problemset/problem/1474

【题意】

【题解】

题目的意思是说,那个块在最左端先出现,就先处理那个块;
每个连通块,处理出最下的行标和最上的行标，以及最左的列标和最右的列表;
把在这个连通块里面的数字做上标记就好(注意不要输出其他连通块的1)

【Number Of WA】

0

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)typedef pair<int,int> pii;
typedef pair<LL,LL> pll;const int tx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int ty[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 510;int a[N][N],n,m,b[N][N],tot,xx,xy,dx,dy;
char s[N];void dfs(int x,int y)
{if (a[x][y]==0) return;a[x][y] = 0,b[x][y] = tot;if (x>dx) dx = x;if (x<xx) xx = x;if (y>dy) dy = y;if (y<xy) xy = y;rep1(i,1,4)dfs(x+tx[i],y+ty[i]);
}void out()
{cout << dx-xx+1<<' '<<dy-xy+1<<endl;rep1(i,xx,dx){rep1(j,xy,dy)if (b[i][j]==tot)cout<<1;elsecout<<0;cout << endl;}
}int main()
{//freopen("F:\\rush.txt","r",stdin);ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not usecin >> n >> m;rep1(i,1,n){cin >> (s+1);rep1(j,1,m)if (s[j]=='1')a[i][j] = 1;elsea[i][j] = 0;}rep1(j,1,m){rep1(i,1,n)if (a[i][j]){tot++;xx = i,xy = j,dx=i,dy=j;dfs(i,j);out();}}return 0;
}