CodeForces 6B President's Office
2016-09-27 20:30
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题意:在一个n行m列的办公室中有一些小的矩形桌子,每个人的桌子都有唯一的颜色与之对应,总统的副手与总统桌子相邻,给出一个图,c代表总统桌子颜色,求总统有多少个副手。
思路:从总统的桌子开始找,相邻的桌子,并将其去掉,没去掉一次人数加一次,写一个去掉桌子的函数。
#include<iostream>
using namespace std;
const int maxx = 101;
char map[maxx][maxx];
char c;
int num,n,m;
int cal[5] = {1,0,-1,0,1};
void clear(int x,int y);
void dfs(int x,int y)
{
if(map[x][y]==c)
{
for(int i=0;i<4;i++)// x+1,y x,y+1 x-1,y x,y-1
{
int x1=x+cal[i],y1=y-cal[i+1];
if(map[x1][y1]!='.'&&x1>=0&&x1<n&&y1>=0&&y1<m)
{
map[x][y]='.';
dfs(x1,y1);
}
}
}
else if(map[x][y]!='.')
{
clear(x,y);
num++;
}
}
void clear(int x,int y)
{
char b = map[x][y];
map[x][y]='.';
for(int i=0;i<4;i++)
{
int x1=x+cal[i],y1=y-cal[i+1];
if(map[x1][y1]==b&&x1>=0&&x1<n&&y1>=0&&y1<m)
{
clear(x1,y1);
}
}
}
int main()
{
int x,y;
while(cin>>n>>m)
{
cin>>c;
num=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
cin>>map[i][j];
}
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(map[i][j]==c)
{
x=i;
y=j;
break;
}
}
dfs(x,y);
cout<<num<<endl;
}
}
思路:从总统的桌子开始找,相邻的桌子,并将其去掉,没去掉一次人数加一次,写一个去掉桌子的函数。
#include<iostream>
using namespace std;
const int maxx = 101;
char map[maxx][maxx];
char c;
int num,n,m;
int cal[5] = {1,0,-1,0,1};
void clear(int x,int y);
void dfs(int x,int y)
{
if(map[x][y]==c)
{
for(int i=0;i<4;i++)// x+1,y x,y+1 x-1,y x,y-1
{
int x1=x+cal[i],y1=y-cal[i+1];
if(map[x1][y1]!='.'&&x1>=0&&x1<n&&y1>=0&&y1<m)
{
map[x][y]='.';
dfs(x1,y1);
}
}
}
else if(map[x][y]!='.')
{
clear(x,y);
num++;
}
}
void clear(int x,int y)
{
char b = map[x][y];
map[x][y]='.';
for(int i=0;i<4;i++)
{
int x1=x+cal[i],y1=y-cal[i+1];
if(map[x1][y1]==b&&x1>=0&&x1<n&&y1>=0&&y1<m)
{
clear(x1,y1);
}
}
}
int main()
{
int x,y;
while(cin>>n>>m)
{
cin>>c;
num=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
cin>>map[i][j];
}
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
if(map[i][j]==c)
{
x=i;
y=j;
break;
}
}
dfs(x,y);
cout<<num<<endl;
}
}
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