The Pilots Brothers' refrigerator Poj2965
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 20326 | Accepted: 7831 | Special Judge | ||
Description
The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator.
There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerator is open only when all handles are open. The handles are represented as a matrix 4х4. You can change the state of a handle in any location [i, j] (1 ≤ i, j ≤ 4). However, this also changes states of all handles in row i and all handles in column j.
The task is to determine the minimum number of handle switching necessary to open the refrigerator.
Input
The input contains four lines. Each of the four lines contains four characters describing the initial state of appropriate handles. A symbol “+” means that the handle is in closed state, whereas the symbol “−” means “open”. At least one of the handles is initially closed.
Output
The first line of the input contains N – the minimum number of switching. The rest N lines describe switching sequence. Each of the lines contains a row number and a column number of the matrix separated by one or more spaces. If there are several solutions, you may give any one of them.
Sample Input
-+-- ---- ---- -+--
Sample Output
6 1 1 1 3 1 4 4 1 4 3 4 4
第一次写博客 :)(纪念一下)。
此题为BFS纯收索。用的一位数组，关键是将对一维数组的操作间接地对二维数组进行变动。
#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int map[20];
struct node
{
int x,ans,a[20],vis[20];//有输出进行改动的坐标，故将他们分别的标记。
} q[100010];
void bfs()
{
int i,j,k,s=0,e=0;
node f1,f2;
for(i=0; i<16; i++)
{
f1.a[i]=map[i];
f1.vis[i]=0;
}
f1.x=f1.ans=0;
q[s++]=f1;
while(s>e)
{
f1=q[e++];
int ss=0;
for(i=0; i<16; i++)
ss+=f1.a[i];
if(ss==0)
{
cout<<f1.ans<<endl;
for(i=0; i<16; i++)
if(f1.vis[i])
cout<<i/4+1<<" "<<i%4+1<<endl;// /4是对行，%4对列
return ;
}
for(i=0; i<16; i++)
{
if(f1.x<=i)//一定不要少了，此为对移动次数的控制即走过一次便少遍历一次
{
f2=f1;
for(j=i-i%4; j<i+4-i%4; j++)//行 i-i%4即以i=》4--7来说j都是从4--7实现的对同行的改变
f2.a[j]=1-f2.a[j];
for(j=i%4; j<16; j=j+4)
f2.a[j]=1-f2.a[j];
f2.a[i]=1-f2.a[i];///注意下标是i不是j
f2.x=i+1;
f2.ans=f1.ans+1;
f2.vis[i]=1;
q[s++]=f2;
}
}
}
puts("Impossible");
return ;
}
char s[4];
int main()
{
int n,m,i,j,k;
for(i=0; i<4; i++)
{
cin>>s;
for(j=0; j<4; j++)
{
if(s[j]=='-')
{
map[i*4+j]=0;
}
else
map[4*i+j ]=1;
}
}
bfs();
return 0;
}
后来再做的时候还是有些不足
1：在限制放16个棋子的时候居然写成了f2.tmp++
2:要使用手动模拟队列，快！
#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cstdlib> #include<string> #include<queue> #include<map> #define L1 long long #define L2 __int64 #define inf 0x3f3f3f3f #pragma comment(linker, "/STACK:1024000000,1024000000") using namespace std; //const int m1=1001000; //const int m2=1010; //int head[m1],vex[m1],arr[m1],cnt; bool vis[17]; int cnt; char s1[5][5]; struct node { int tmp,li; bool vis[20]; char s[17]; }q[100010]; void bfs() { int i,j,s,e; s=e=0; node f1,f2; f1.tmp=f1.li=0; cnt=0; for(i=0; i<4; i++) for(j=0; j<4; j++) { f1.vis[cnt]=false; f1.s[cnt++]=s1[i][j]; } f1.s[cnt]='\0'; q[s++]=f1; //cout<<f1.s<<endl; int t; while(s>e) { f1=q[e++]; bool bj=false; for(i=1;i<16;i++) { if(f1.s[i]==f1.s[i-1]&&f1.s[i]=='-') { } else { bj=true; break; } } //cout<<f1.s<<endl; if(!bj) { printf("%d\n",f1.tmp); for(i=0;i<16;i++) { if(f1.vis[i]) { printf("%d %d\n",i/4+1,i%4+1); } } return ; } for(i=0; i<16; i++) { if(f1.li<=i) { f2=f1; t=i/4; for(j=t*4;j<t*4+4;j++) { if(f2.s[j]=='+') f2.s[j]='-'; else f2.s[j]='+'; } t=i%4; for( j=t;j<16;j+=4) { if(f2.s[j]=='+') f2.s[j]='-'; else f2.s[j]='+'; } if(f2.s[i]=='+') f2.s[i]='-'; else f2.s[i]='+'; f2.li=i+1; f2.tmp++; f2.vis[i]=true; q[s++]=f2; } } } return ; } int main() { int n,m,i,j,k; for(i=0; i<4; i++) scanf("%s",s1[i]); bfs(); return 0; }