red and back

来源:互联网 发布:江西软件先锋学院 编辑:程序博客网 时间:2024/06/05 11:07
#include<iostream>#include<cstring>#include<queue>#define maxn 205using namespace std;int K[maxn],vis[maxn];void bfs(int s,int e){    queue<int> Q;    Q.push(s);    vis[s]=1;    int t;    while(!Q.empty())    {        int next;        t=Q.front();        Q.pop();        if(t==e) break;        next=t+K[t];        if(next>=1&&next<=e&&vis[next]==0)        {            Q.push(next);            vis[next]=vis[t]+1;        }        next=t-K[t];        if(next>=1&&next<=e&&vis[next]==0)        {            Q.push(next);            vis[next]=vis[t]+1;        }    }    if(t!=e) vis[e]=0;}int main(){    int n,s,e;    while(cin>>n&&n)    {memset(vis,0,sizeof(vis));        cin>>s>>e;        for(int i=1;i<=n;i++)            cin>>K[i];                bfs(s,e);        cout<<vis[e]-1<<endl;    }    return 0;}
# include<iostream>#include<cstdio># include<cstring># include<queue>using namespace std;const int maxn=25;int dx[]={0,0,1,-1};int dy[]={1,-1,0,0}; int n,m,sum,sx,sy; struct point{    int x,y; };char a[maxn][maxn];bool  vis[maxn][maxn];int bfs(int xx,int yy ){     queue<point> Q;    point in,next,cur;     in.x=xx;     in.y=yy;     a[in.x][in.y]='#';     Q.push(in);    while(!Q.empty())    {        cur=Q.front();        Q.pop();        for(int i=0;i<4;i++)        {              next.x=cur.x+dx[i];               next.y=cur.y+dy[i];           if(next.x>=0&&next.x<m&&next.y>=0&&next.y<n)              if(a[next.x][next.y]!='#')              {                  sum++;                  a[next.x][next.y]='#';                  Q.push(next);              }        }    }    return sum;}int main(){    while(cin>>n>>m,n,m)    {        sum=1;        memset(vis,0,sizeof(vis));        for(int i=0;i<m;i++)            for(int j=0;j<n;j++)              {cin>>a[i][j];                 if(a[i][j]=='@')                  {  sx=i;sy=j;}              }              sum=bfs(sx,sy);              cout<<sum<<endl;    }    return 0;}


Red and Black

 
Problem Description
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.
 

Input
The 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)
 

Output
For 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
4559613
 

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