There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
InputThe input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)
OutputFor each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
Sample Input
6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
思路：这道题我参考了书本然后用了bfs，在这个矩阵中，点只能往四个方向走，分别是上下左右，所以设置数组dir来实现坐标的变化，check则是判断这个点有没有超出矩阵的大小，结构体就是为了存储这个点的坐标而建立的。
然后是bfs函数，因为题目出发的点也算一个，所以sum初始值为1，然后建立队列，把第一个点压入队列，开始循环一直到队列为空（也就是把矩阵都访问一遍）。对每一个点都进行四次循环，如果周围的点中有可以走的点，num加1，然后把这个点标记为不可走，并把这个点压入队列。如此循环往复，直到队列为空。
ac代码：
#include <iostream> #include <queue> using namespace std; char room[23][23]; int dir[4][2]={{-1,0},{0,-1},{1,0},{0,1}}; int wx,hy,num; #define check(x,y)(x<wx&&x>=0&&y>=0&&y<hy) struct node{int x,y;}; void bfs(int dx,int dy){ num=1; queue<node>q; node start,next; start.x=dx; start.y=dy; q.push(start); while(!q.empty()){ start=q.front(); q.pop(); for(int i=0;i<4;i++){ next.x=start.x+dir[i][0]; next.y=start.y+dir[i][1]; if(check(next.x,next.y)&&room[next.x][next.y]=='.'){ room[next.x][next.y]='#'; num++; q.push(next); } } } } int main(int argc, const char * argv[]) { int x,y,dx,dy; while(cin>>wx>>hy){ if(wx==0&&hy==0){ break; } for(y=0;y<hy;y++){ for(x=0;x<wx;x++){ cin>>room[x][y]; if(room[x][y]=='@'){ dx=x; dy=y; } } } num=0; bfs(dx,dy); cout<<num<<endl; } return 0; }
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.
* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.
If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
Input
Line 1: Two space-separated integers: N and K
Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
Sample Input
5 17
Sample Output
4
Hint
The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.
思路：这道题目也是一道入门的bfs的题目，主要是要额外设置一个标记数组，然后在设置一个步数数组，直接计算走到哪一个数字需要的步数。因为这道题只有三个方向，并且是在一位数组上，所以就直接暴力把每一个压入队列的数字都进行+1，-1和*2操作，然后再把这些也全部压入队列中，然后加入了判断，只要与奶牛位置相同就退出循环输出步数。
ac代码：
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#include <iostream> #include <queue> using namespace std; int bfs(int n,int k){ int flag1[1000001]; int bs[1000001]; queue<int>q; q.push(n); int start=0; if(n==k){ return 0; } while(!q.empty()){ start=q.front(); q.pop(); if(start+1<1000001&&flag1[start+1]!=1){ flag1[start+1]=1; bs[start+1]=bs[start]+1; q.push(start+1); } if(start+1==k) break; if(start-1>=0&&flag1[start-1]!=1){ flag1[start-1]=1; bs[start-1]=bs[start]+1; q.push(start-1); } if(start-1==k) break; if(start*2<1000001&&flag1[start*2]!=1){ flag1[start*2]=1; bs[start*2]=bs[start]+1; q.push(start*2); } if(start*2==k) break; } return bs[k]; } int main(int argc, const char * argv[]) { int N,K; cin>>N>>K; cout<<bfs(N, K)<<endl; return 0; }
Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than 100 decimal digits.
Input
The input file may contain multiple test cases. Each line contains a value of n (1 <= n <= 200). A line containing a zero terminates the input.
Output
For each value of n in the input print a line containing the corresponding value of m. The decimal representation of m must not contain more than 100 digits. If there are multiple solutions for a given value of n, any one of them is acceptable.
Sample Input
2 6 19 0
Sample Output
10 100100100100100100 111111111111111111
思路：这道题要找一个数能够整除题目中给的数，这道题目也只有两个方向，从1开始，要不就是*10，要不就是*10+1，利用bfs遍历输出其中最小的一个就行了
ac代码：
#include <iostream> #include <queue> using namespace std; void bfs(long long n,int m){ queue<long long>q; q.push(n); while(!q.empty()){ long long t=q.front(); q.pop(); if(t%m==0){ cout<<t<<endl; return ; } else {q.push(t*10);q.push(t*10+1);} } } int main(int argc, const char * argv[]) { int k; bool flag1=true; while(flag1){ cin>>k; if(k==0){ flag1=false; break; } if(k!=0){ bfs(1,k); } } return 0; }
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
1733
3733
3739
3779
8779
8179
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
3
思路；这道题目就是将一个质数通过最少的步骤转换成另外一个步数，换一个数字要➕1，然后主要是四个方向的变化，分别为个十百千位的变化，但是要注意的是个位必须是奇数，千位要大于等于1，然后我不会写质数表，就写了一个判断它是不是质数的函数，然后利用bfs就可以解决了
ac代码：
#include <iostream> #include <string.h> #include <cmath> #include <algorithm> #include <queue> using namespace std; #define max 1000000 int bool1[max]; struct node{ int data; int step; }x,y; bool judge(int x)//判断是否为素数// { if(x==1||x==0) return 0; for(int i = 2; i <= sqrt(x); i ++) { if(x % i == 0) return 0; } return 1; } //变化个位：X/10*10+i;变化十位：X/100*100+i*10+X%10;变化百位：X/1000*1000+i*100+i%100;变化千位：i*1000+i%1000;// void bfs(int n,int m){ queue<node>q; memset(bool1,0,sizeof(bool1)); bool1[n]=1; x.data=n; x.step=0; q.push(x); while(!q.empty()){ x=q.front(); q.pop(); if(x.data==m){ cout<<x.step<<endl; return; } //个位// for(int i=1;i<=9;i=i+2){ int k=x.data/10*10+i; if(judge(k)&&!bool1[k]){ bool1[k]=1; y.data=k; y.step=x.step+1; q.push(y); } } //十位// for(int i=0;i<=9;i++){ int t=x.data%10; int k=x.data/100*100 +i*10+t; if(judge(k)&&!bool1[k]){ bool1[k]=1; y.data=k; y.step=x.step+1; q.push(y); } } //百位// for(int i=0;i<=9;i++){ int t=x.data%100; int k=x.data/1000*1000 +i*100+t; if(judge(k)&&!bool1[k]){ bool1[k]=1; y.data=k; y.step=x.step+1; q.push(y); } } //千位// for(int i=1;i<=9;i++){ int t=x.data%1000; int k=i*1000+t; if(judge(k)&&!bool1[k]){ bool1[k]=1; y.data=k; y.step=x.step+1; q.push(y); } } } return; } int main(int argc, const char * argv[]) { int num; int k,l; cin>>num; while(num--){ cin>>k>>l; bfs(k,l); } return 0; }
Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.
Given a diagram of Farmer John's field, determine how many ponds he has.
Given a diagram of Farmer John's field, determine how many ponds he has.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.
Output
* Line 1: The number of ponds in Farmer John's field.
Sample Input
10 12 W........WW. .WWW.....WWW ....WW...WW. .........WW. .........W.. ..W......W.. .W.W.....WW. W.W.W.....W. .W.W......W. ..W.......W.
Sample Output
3
Hint
OUTPUT DETAILS:
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
There are three ponds: one in the upper left, one in the lower left,and one along the right side.
思路：这道题就是求一个区域内最大连通子图的数量，然后这道题需要八个方向的变化，因为要判断这个点周围八个区域是不是相应的联通区域，然后利用dfs递归的方式将读到这个点的所有连通区域变成不可以走的范围，然后统计读到几个点就是连通区域的数量
ac代码：
#include <iostream> #define max 107; using namespace std; int dir[8][2]={{-1,0},{1,0},{0,-1},{0,1},{-1,-1},{1,1},{1,-1},{-1,1}}; char a[107][107]; int n,m; void dfs(int x,int y){ a[x][y]='.'; for(int i=0;i<8;i++){ int x1,y1; x1 = x+dir[i][0]; y1 = y+dir[i][1]; if(x1>n-1||x1<0||y1<0||y1>m-1){ continue; } if(a[x1][y1]=='.'){ continue; } dfs(x1,y1); } } int main(int argc, const char * argv[]){ while(cin>>n>>m){ int sum=0; if(n==0||m==0){ break; } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ cin>>a[i][j]; } } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ if(a[i][j]=='W'){ dfs(i, j); cout<<i<<j<<endl; sum=sum+1; } } } cout<<sum; } return 0; }