POJ 3278 Catch That Cow
用一个vis数组标记一下每个点是否被访问过,否则TLE或者MLE…
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int N=1e5;
int a,b,x,nx,cnt,vis[N+10];
struct node
{
int x,cnt;
};
queue<node>q;
int main()
{
ios::sync_with_stdio(false);
cin>>a>>b;
if(a>=b){printf("%d\n",a-b);return 0;}
q.push({a,0});
vis[a]=1;
while(!q.empty())
{
node tmp=q.front();q.pop();
x=tmp.x;
cnt=tmp.cnt;
if(x==b){printf("%d\n",cnt);break;}
nx=x+1;
if(nx>=0&&nx<=N&&!vis[nx])
{
vis[nx]=1;
q.push({nx,cnt+1});
}
nx=x-1;
if(nx>=0&&nx<=N&&!vis[nx])
{
vis[nx]=1;
q.push({nx,cnt+1});
}
nx=x*2;
if(nx>=0&&nx<=N&&!vis[nx])
{
vis[nx]=1;
q.push({nx,cnt+1});
}
}
return 0;
}
《算法竞赛进阶指南》题目练习 CH2501 矩阵距离
从1开始搜索,搜索到0时记录最短步数,保存到ans数组中。
#include <bits/stdc++.h>
using namespace std;
const int N=1010;
char a[N][N];
int n,m,ans[N][N],vis[N][N];
int dir[4][2]={0,1,0,-1,1,0,-1,0};
struct node
{
int x,y,cnt;
};
queue<node>q;
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m;
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
{
cin>>a[i][j];
if(a[i][j]=='1')q.push({i,j,0}),ans[i][j]=0,vis[i][j]=1;
}
while(!q.empty())
{
node tmp=q.front();q.pop();
int x=tmp.x,y=tmp.y,cnt=tmp.cnt;
for(int i=0;i<4;i++)
{
int nx=x+dir[i][0];
int ny=y+dir[i][1];
if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&!vis[nx][ny])
{
vis[nx][ny]=1;
ans[nx][ny]=cnt+1;
q.push({nx,ny,cnt+1});
}
}
}
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
j==m?printf("%d\n",ans[i][j]):printf("%d ",ans[i][j]);
return 0;
}