标签：des blog http io os java ar strong for
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 31195 | Accepted: 10668 |
Description
Background Input
Output
Sample Input
3 1 1 2 3 4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4
实际用时 20min
情况:CCCCA 注意java和胡乱改动
注意点:1 组间空行但最后一组没有 2 字典序
#include <cstdio>
#include <cstring>
using namespace std;
int n,m;
typedef unsigned long long ull;
bool used[8][8];
char heap[64][3];
const int dx[8]={-2,-2,-1,-1,1,1,2,2},dy[8]={-1,1,-2,2,-2,2,-1,1};
bool judge(int x,int y){
if(x>=0&&x<n&&y>=0&&y<m)return true;
return false;
}
bool dfs(int x,int y,int cnt){
used[x][y]=true;
heap[cnt][0]=x+‘A‘;
heap[cnt++][1]=y+‘1‘;
if(cnt==n*m)return true;
for(int i=0;i<8;i++){
int tx=x+dx[i],ty=y+dy[i];
if(judge(tx,ty)&&!used[tx][ty]){
if(dfs(tx,ty,cnt))return true;
}
}
used[x][y]=false;
return false;
}
int main(){
int T;scanf("%d",&T);
for(int ti=1;ti<=T;ti++){
scanf("%d%d",&m,&n);
memset(used,0,sizeof(used));
bool fl=dfs(0,0,0);
printf("Scenario #%d:\n",ti);
if(fl){
for(int i=0;i<n*m;i++){
printf("%s",heap[i]);
}
puts("");
}
else {
puts("impossible");
}
if(ti<T)puts("");
}
return 0;
}
快速切题 poj2488 A Knight's Journey
标签：des blog http io os java ar strong for
原文地址：http://www.cnblogs.com/xuesu/p/3992373.html
