题目
地图连通性测试

方法一:常规深搜
cpp
#include <iostream>
#include <iomanip>
using namespace std;
int n;
int a[1010][1010];
int cnt;
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};
void dfs(int,int);
int main()
{
cin>>n;
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
cin>>a[i][j];
}
}
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
if(a[i][j]==1)
{
if(cnt>0)
{
cout<<"NO";
return 0;
}
cnt++;
dfs(i,j);
}
}
}
cout<<"YES";
return 0;
}
void dfs(int x,int y)
{
a[x][y] = 0;
for(int i = 0;i<4;i++)
{
int tx = x+dx[i];
int ty = y+dy[i];
if(tx>=1&&tx<=n&&ty>=1&&ty<=n&&a[tx][ty]==1)
{
dfs(tx,ty);
}
}
}
方法二:用栈实现深搜
cpp
#include <bits/stdc++.h>
using namespace std;
struct node
{
int x,y;
node(){};
node(int aaaaa,int bbbbb)
{
x = aaaaa;
y = bbbbb;
}
};
stack<node> b;
int n;
int a[1010][1010];
int cnt;
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};
void dfs(int,int);
int main()
{
cin>>n;
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
cin>>a[i][j];
}
}
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
if(a[i][j]==1)
{
if(cnt>0)
{
cout<<"NO";
return 0;
}
cnt++;
dfs(i,j);
}
}
}
cout<<"YES";
return 0;
}
void dfs(int x1,int y1)
{
b.push({x1,y1});
while(b.size()!=0)
{
node p = b.top();
b.pop();
if(a[p.x][p.y]==0) continue;
a[p.x][p.y] = 0;
for(int i = 3;i>=0;i--)
{
int tx = p.x+dx[i];
int ty = p.y+dy[i];
if(tx>=1&&tx<=n&&ty>=1&&ty<=n)
{
b.push({tx,ty});
}
}
}
}
泳池

方法一:常规深搜
cpp
#include <bits/stdc++.h>
using namespace std;
int n;
int a[1010][1010];
int cnt;
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};
bool dfs(int,int);
int main()
{
cin>>n;
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
cin>>a[i][j];
}
}
cnt = a[n][n];
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
a[i][j] = max(0,a[i][j]-a[n][n]);
}
}
while(true)
{
if(dfs(1,1)==true)
{
cout<<cnt;
return 0;
}
cnt++;
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
a[i][j] = max(a[i][j]-1,0);
}
}
}
return 0;
}
bool dfs(int x,int y)
{
if(x==n&&y==n) return true;
bool f = false;
for(int i = 0;i<3;i++)
{
int tx = x+dx[i];
int ty = y+dy[i];
if(tx>=1&&tx<=n&&ty>=1&&ty<=n&&a[tx][ty]==0)
{
a[tx][ty] = -1;
f = dfs(tx,ty);
a[tx][ty] = 0;
if(f==true) return true;
}
}
return false;
}
方法二:用栈实现深搜
cpp
#include <bits/stdc++.h>
using namespace std;
struct node
{
int x,y;
node(){};
node(int aaaaa,int bbbbb)
{
x = aaaaa;
y = bbbbb;
}
};
stack<node> b;
int n;
int a[1010][1010];
int cnt;
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};
bool dfs(int,int);
int main()
{
cin>>n;
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
cin>>a[i][j];
}
}
cnt = a[n][n];
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
a[i][j] = max(a[i][j]-a[n][n],0);
}
}
while(true)
{
if(dfs(1,1)==true)
{
cout<<cnt;
return 0;
}
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
a[i][j] = max(a[i][j]-1,0);
}
}
cnt++;
}
return 0;
}
bool dfs(int x1,int y1)
{
b.push({x1,y1});
while(b.size()!=0)
{
if(b.top().x==n&&b.top().y==n) return true;
node p = b.top();
b.pop();
if(a[p.x][p.y]!=0) continue;
a[p.x][p.y] = -1;
for(int i = 3;i>=0;i--)
{
int tx = p.x+dx[i];
int ty = p.y+dy[i];
if(tx>=1&&tx<=n&&ty>=1&&ty<=n)
{
b.push({tx,ty});
if(tx==n&&ty==n) return true;
}
}
}
return false;
}
晶矿的个数

方法一:常规深搜
cpp
#include <bits/stdc++.h>
using namespace std;
int n;
char a[1010][1010];
int cntr,cntb;
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};
void dfs(int,int,char);
int main()
{
int t;
cin>>t;
while(t--)
{
cin>>n;
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
cin>>a[i][j];
}
}
cntr = 0;
cntb = 0;
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
if(a[i][j]=='r'||a[i][j]=='b')
{
if(a[i][j]=='r') cntr++;
else cntb++;
dfs(i,j,a[i][j]);
}
}
}
cout<<cntr<<" "<<cntb<<endl;
}
return 0;
}
void dfs(int x,int y,char ch)
{
a[x][y] = '#';
for(int i = 0;i<=3;i++)
{
int tx = x+dx[i];
int ty = y+dy[i];
if(tx>=1&&tx<=n&&ty>=1&&ty<=n&&a[tx][ty]==ch)
{
dfs(tx,ty,ch);
}
}
}
方法二:用栈实现深搜
cpp
#include <bits/stdc++.h>
using namespace std;
struct node
{
int x,y;
node(){};
node(int aaaaa,int bbbbb)
{
x = aaaaa;
y = bbbbb;
}
};stack<node> b;
int n;
char a[1010][1010];
int cntr,cntb;
int dx[] = {0,1,0,-1};
int dy[] = {1,0,-1,0};
void dfs(int,int,char);
int main()
{
int t;
cin>>t;
while(t--)
{
cin>>n;
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
cin>>a[i][j];
}
}
cntr = 0;
cntb = 0;
for(int i = 1;i<=n;i++)
{
for(int j = 1;j<=n;j++)
{
if(a[i][j]=='r'||a[i][j]=='b')
{
if(a[i][j]=='r') cntr++;
else cntb++;
dfs(i,j,a[i][j]);
}
}
}
cout<<cntr<<" "<<cntb<<endl;
}
return 0;
}
void dfs(int x1,int y1,char ch)
{
b.push({x1,y1});
while(b.size()>0)
{
node p = b.top();
b.pop();
a[p.x][p.y] = '#';
for(int i = 3;i>=0;i--)
{
int tx = p.x+dx[i];
int ty = p.y+dy[i];
if(tx>=1&&tx<=n&&ty>=1&&ty<=n&&a[tx][ty]==ch)
{
b.push({tx,ty});
}
}
}
}