Lake Counting（dfs） + Catch That Cow（bfs）Lake CountingCatch That Cow

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概述

【一】

Lake Counting

TimeLimit:1000MS MemoryLimit:65536K

64-bit integer IO format:%lld

Problem Description

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.

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.

Output

* Line 1: The number of ponds in Farmer John's field.

SampleInput

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

SampleOutput

【题解】

从任意的'W'开始将周围的八个方块包括自己九个方块变成'.'，并把八个方块中为'W'的方块的位置做重复操作，记录这样变化的次数并输出。

【代码】

#include<stdio.h>
char a[105][105];
int n,m;
void dfs(int x,int y)
{
for(int i=-1;i<=1;i++)
for(int j=-1;j<=1;j++)
{
int xx=x+i,yy=y+j;
if(xx>=0&&xx<n&&yy>=0&&yy<m&&a[xx][yy]=='W')
{
a[xx][yy]='.';
dfs(xx,yy);
}
}
}
main()
{
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%s",a[i]);
int ans=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
if(a[i][j]=='W')
{
dfs(i,j);
ans++;
}
printf("%dn",ans);
}

【二】

Catch That Cow

TimeLimit:2000MS MemoryLimit:32768KB

64-bit integer IO format:%I64d

Problem Description

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?

多组数据

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.

SampleInput

5 17

SampleOutput

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.

【题解】

显然如果n>=k 那么只能采取x-1的方式到达k。总共有三种走法分别为x+1,x-1,2*x，判断范围和是否走过，直接bfs即可。

【代码】

#include<bits/stdc++.h>
using namespace std;
int n,k;
int vis[100005];
struct p{
int x,c;
}st,ed;
int bfs()
{
queue <p> q;
while(!q.empty())
q.pop();
st.x=n,st.c=0;
q.push(st);
while(!q.empty())
{
st=q.front();
q.pop();
for(int i=1;i<=3;i++)
{
ed.c=st.c+1;
if(i==1)
ed.x=st.x+1;
else if(i==2)
ed.x=st.x-1;
else
ed.x=st.x*2;
if(ed.x==k)
return ed.c;
if(ed.x>=0&&ed.x<=100000&&vis[ed.x]==0)
{
vis[ed.x]=1;
q.push(ed);
}
}
}
}
main()
{
while(~scanf("%d%d",&n,&k))
{
if(n>=k)
printf("%dn",n-k);
else
{
memset(vis,0,sizeof(vis));
printf("%dn",bfs());
}
}
}