基础BFS 邻接矩阵Flood fill 算法题目总结

很基础的一类BFS题，个人觉得BFS在求最短、最近、最快等题目时比DFS还是有较大优势的

Poj 1979 Red and Black

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int N=25;
struct Point
{
	int x,y;
};

int graph

;
bool visit

;
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};
int W,H,u,v;

int OK (int x,int y)
{
	return x>=0 && x<W && y>=0 && y<H;
}

int BFS ()
{
	int ans=1;
	Point top,t;
    queue <Point> Q;
	top.x=u;
	top.y=v;
	Q.push(top);
	visit[top.x][top.y]=true;
	while (Q.empty()==false)
	{
		top=Q.front();
		Q.pop();
		for (int i=0;i<4;i++)
		{
			t.x=top.x+dx[i];
			t.y=top.y+dy[i];
			if (OK(t.x,t.y) && graph[t.x][t.y] && !visit[t.x][t.y])
			{
				visit[t.x][t.y]=true;
				Q.push(t);
				ans++;
			}
		}
	}
	return ans;
}

int main ()
{
	while (scanf("%d%d",&W,&H) , (W || H))
	{
        char str
;
        memset(visit,false,sizeof(visit));
        memset(graph,0,sizeof(graph));
        for (int i=0;i<H;i++)
        {
            scanf("%s",str);
            for (int j=0;j<W;j++)
            {
                if (str[j] == '.')
                    graph[j][i]=1;
                if (str[j] == '@')
                {
                    graph[j][i]=1;   //u,v存起点坐标
                    u=j;
                    v=i;
                }
            }
        }
        printf("%d\n",BFS());
    }
	return 0;
}

Poj 2243 Knight Moves

题意：要求用象棋中马跳动方式到达终点，求马需要跳动的最少次数。

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int N=10;

struct Point
{
	int x,y,step;
};

int X,Y,n,u,v;
bool visit

;
int dx[8] = {1,2,2,1,-1,-2,-2,-1};
int dy[8] = {2,1,-1,-2,-2,-1,1,2};

int OK (int x,int y)
{
	return x>=0 && x<=7 && y>=0 && y<=7 && visit[x][y]==false;
}

int BFS ()
{
	Point top,t;
	queue <Point> Q;
	top.step=0;
	top.x=u;
	top.y=v;
	Q.push(top);
	visit[top.x][top.y]=true;
	while (Q.empty()==false)
	{
		top=Q.front();
		Q.pop();
		if (top.x==X && top.y==Y)
			break;
		for (int i=0;i<8;i++)
		{
			t.x=top.x+dx[i];
			t.y=top.y+dy[i];
			t.step=top.step+1;
			if (OK(t.x,t.y) && !visit[t.x][t.y])
			{
				visit[t.x][t.y]=true;
				Q.push(t);
			}
		}
	}
	return top.step;
}

int main ()
{
	char ch1[4],ch2[4];
	while (~scanf("%s%s",&ch1,&ch2))
	{
		memset(visit,false,sizeof(visit));
		u=ch1[0]-'a';
		v=ch1[1]-'1';
		X=ch2[0]-'a';
		Y=ch2[1]-'1';
		printf("To get from %s to %s takes %d knight moves.\n",ch1,ch2,BFS());
	}
	return 0;
}

Poj 1915 Knight Moves

题意：输入n代表有个n*n的棋盘，输入开始位置的坐标和结束位置的坐标，

问一个骑士朝棋盘的八个方向走马字步，从开始坐标到结束坐标可以经过多少步。

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int N=305;

struct Point
{
	int x,y,step;
};

int X,Y,n,u,v;
bool visit

;
int dx[8] = {1,2,2,1,-1,-2,-2,-1};
int dy[8] = {2,1,-1,-2,-2,-1,1,2};
queue <Point> Q;

int OK (int x,int y)
{
	return x>=0 && x<=n-1 && y>=0 && y<=n-1 && visit[x][y]==false;
}

int BFS ()
{
	Point top,t;
	int i;
	top.step=0;
	top.x=u;
	top.y=v;
	Q.push(top);
	visit[top.x][top.y]=true;
	while (Q.empty()==false)
	{
		top=Q.front();
		Q.pop();
		if (top.x==X && top.y==Y)
			break;
		for (i=0;i<8;i++)
		{
			t.x=top.x+dx[i];
			t.y=top.y+dy[i];
			t.step=top.step+1;
			if (OK(t.x,t.y))
			{
				visit[t.x][t.y]=true;
				Q.push(t);
			}
		}
	}
	return top.step;
}

int main ()
{
	int T;
	scanf("%d",&T);
	while (T--)
	{
		memset(visit,false,sizeof(visit));
		while (Q.empty()==false)
            Q.pop();
		scanf("%d%d%d%d%d",&n,&X,&Y,&u,&v);
		printf("%d\n",BFS());
	}
	return 0;
}

LightOJ 1012 Guilty Prince

题意:给两个数,w,h,表示给一个 高h,宽w的矩阵,

‘#’表示不能走,‘.’表示能走,‘@’表示起始点,求从起始点出发能访问多少个点。

#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;

const int N=25;
char g

;
bool visit

;
int w,h;

int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};

struct Node
{
	int x,y;
}temp,cur,s;

bool OK (Node a)
{
	if (visit[a.x][a.y]==false && g[a.x][a.y]!='#')
		return true;
	return false;
}

int BFS ()
{
	int cnt=0;
	queue<Node> q;
	q.push(s);
	while (!q.empty())
	{
		cur = q.front();
		q.pop();
		for (int i=0;i<4;i++)
		{
			temp.x=cur.x+dx[i];
			temp.y=cur.y+dy[i];
			if (OK(temp))
			{
				visit[temp.x][temp.y]=true;
				q.push(temp);
				cnt++;
			}
		}
	}
	return cnt;
}

int main ()
{
	int T;
	scanf("%d",&T);
	for (int Cas=1;Cas<=T;Cas++)
	{
		scanf("%d%d",&w,&h);
		memset(g,'#',sizeof(g));
		memset(visit,false,sizeof(visit));
		for (int i=1;i<=h;i++)
		{
			scanf("%s",&g[i][1]);
			g[i][w+1]='#';
			for (int j=1;j<=w;j++)
				if (g[i][j]=='@')
				{
					s.x=i;
					s.y=j;
					visit[s.x][s.y]=true;
				}
		}
		printf("Case %d: %d\n",Cas,BFS()+1);
	}
	return 0;
}

Poj 3626 Mud Puddles

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int N=1005;

struct Point
{
	int x,y,step;
};

int a

,X,Y,n;
bool visit

;
int dx[]={1,-1,0,0};
int dy[]={0,0,1,-1};

int OK (int x,int y)
{
	return x>=0 && x<=1000 && y>=0 && y<=1000 && visit[x][y]==false;
}

int BFS ()
{
	Point top,t;
	queue <Point> Q;
	top.step=0;
	top.x=top.y=500;
	Q.push(top);
	visit[top.x][top.y]=true;
	while (Q.empty()==false)
	{
		top=Q.front();
		Q.pop();
		if (top.x==X && top.y==Y)
			break;
		for (int i=0;i<4;i++)
		{
			t.x=top.x+dx[i];
			t.y=top.y+dy[i];
			t.step=top.step+1;
			if (OK(t.x,t.y))
			{
				visit[t.x][t.y]=true;
				Q.push(t);
			}
		}
	}
	return top.step;
}

int main ()
{
	while (~scanf("%d%d%d",&X,&Y,&n))
	{
		int i,x,y;
		memset(visit,false,sizeof(visit));
		for (i=1;i<=n;i++)
			scanf("%d%d",&x,&y),visit[x+500][y+500]=true;
		X+=500,Y+=500;
		printf("%d\n",BFS());
	}
	return 0;
}

Poj 2251 Dungeon Master

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int N=50;

struct Point
{
	int x,y,z,step;
};

int visit

,X,Y,Z,x,y,z;
int L,R,C;
int dx[]={1,-1,0,0,0,0};
int dy[]={0,0,1,-1,0,0};
int dz[]={0,0,0,0,1,-1};

int OK (int x,int y,int z)
{
	return x>=1 && x<=R && y>=1 && y<=C && z>=1 && z<=L;
}

void Input ()
{
	char str
;
	memset(visit,true,sizeof(visit));
	for (int k=1;k<=L;k++) for (int i=1;i<=R;i++)
    {
        scanf("%s",str+1);
        for (int j=1;j<=C;j++)
        {
            if (str[j]=='.')
                visit[i][j][k]=false;
            else if(str[j]=='S')
				x=i,y=j,z=k;
            else if(str[j]=='E')
            {
                X=i,Y=j,Z=k;
                visit[i][j][k]=false;
            }
        }
    }
}

int BFS ()
{
	Point top,t;
	queue <Point> Q;
	int time=9999;
	top.step=0;
	top.x=x;
	top.y=y;
	top.z=z;
	Q.push(top);
	visit[top.x][top.y][top.z]=true;
	while (Q.empty()==false)
	{
		top=Q.front();
		Q.pop();
		if (top.x==X && top.y==Y && top.z==Z)
		{
			time=top.step;
			break;
		}
		for (int i=0;i<6;i++)
		{
			t.x=top.x+dx[i];
			t.y=top.y+dy[i];
			t.z=top.z+dz[i];
			t.step=top.step+1;
			if (OK(t.x,t.y,t.z) && visit[t.x][t.y][t.z]==false)
			{
				visit[t.x][t.y][t.z]=true;
				Q.push(t);
			}
		}
	}
	return time;
}

int main ()
{
	while (scanf("%d%d%d",&L,&R,&C) && (L||R||C))
	{
		Input();
		int ans=BFS();
		if (ans==9999)
			printf("Trapped!\n");
		else
			printf("Escaped in %d minute(s).\n",ans);
	}
	return 0;
}