nyoj1038 纸牌游戏

南阳oj第1038题:纸牌游戏
Accept代码:

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#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct Pai
{
char num;
char huase;
};
int comp(const void *a,const void *b)
{
struct Pai *c=(Pai *)a;
struct Pai *d=(Pai *)b;
char cc=(c->huase=='H'?'U':c->huase);
char dd=(d->huase=='H'?'U':d->huase);
char ccc=c->num;
char ddd=d->num;
ccc=(ccc=='A'?'Z':ccc);
ccc=(ccc=='K'?'Y':ccc);
ccc=(ccc=='T'?'B':ccc);
ddd=(ddd=='A'?'Z':ddd);
ddd=(ddd=='K'?'Y':ddd);
ddd=(ddd=='T'?'B':ddd);
if(cc!=dd)
return cc-dd;
else
return ccc-ddd;
}
int main()
{
//freopen("d:\\data\\1.txt","r",stdin);
//freopen("d:\\data\\2.txt","w",stdout);
struct Pai p[4][14];
char s1[150],s2[60];
char person;//先拿牌的人
int yu;
int i,j,t,k;
char shuxu[4][6]={"East","South","West","North"};
while(~scanf("%c",&person))
{
getchar();
scanf("%s",s1);
scanf("%s",s2);
getchar();
strcat(s1,s2);
switch(person)
{
case 'E':yu=0;break;
case 'S':yu=1;break;
case 'W':yu=2;break;
case 'N':yu=3;break;
default :printf("ohmygod\n");
}
for(i=0;i<4;i++)
{
t=0;
k=(i+yu)%4;
for(j=i*2;j<=103;j+=8)
{
p[k][t].num=s1[j];
p[k][t].huase=s1[j+1];
t++;
}
p[k][t].num='\0';
p[k][t].huase='\0';
}
for(i=0;i<4;i++)
{
printf("%s player:\n",shuxu[i]);
qsort(p[i],13,sizeof(p[i][0]),comp);
for(j=0;j<13;j++)
printf("+---+");
printf("\n");
for(j=0;j<13;j++)
printf("|%c %c|",p[i][j].num,p[i][j].num);
printf("\n");
for(j=0;j<13;j++)
printf("| %c |",p[i][j].huase);
printf("\n");
for(j=0;j<13;j++)
printf("|%c %c|",p[i][j].num,p[i][j].num);
printf("\n");
for(j=0;j<13;j++)
printf("+---+");
printf("\n");
}
printf("\n");
}
return 0;
}

nyoj58 最少步数

这题明显是简单的DFS.
Accept代码:

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#include<stdio.h>
int p[9][9]={
1,1,1,1,1,1,1,1,1,
1,0,0,1,0,0,1,0,1,
1,0,0,1,1,0,0,0,1,
1,0,1,0,1,1,0,1,1,
1,0,0,0,0,1,0,0,1,
1,1,0,1,0,1,0,0,1,
1,1,0,1,0,1,0,0,1,
1,1,0,1,0,0,0,0,1,
1,1,1,1,1,1,1,1,1};
int ans,cur;
int x1,x2,y1,y2;
int main()
{
void dfs(int ,int,int);
int N;
scanf("%d",&N);
while(N--)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
p[x1][y1]=1;
ans=100;
dfs(x1,y1,0);
printf("%d\n",ans);
}
return 0;
}
void dfs(int x1,int y1,int cur)
{
if(x1==x2 && y1==y2)
{
ans=cur>ans?ans:cur;
}
if(x1<0 || x1>8 || y1<0 || y1>8)
return ;
if(!p[x1+1][y1]){ p[x1][y1]=1; dfs(x1+1,y1,cur+1); p[x1][y1]=0; }
if(!p[x1][y1+1]){ p[x1][y1]=1; dfs(x1,y1+1,cur+1); p[x1][y1]=0; }
if(!p[x1-1][y1]){ p[x1][y1]=1; dfs(x1-1,y1,cur+1); p[x1][y1]=0; }
if(!p[x1][y1-1]){ p[x1][y1]=1; dfs(x1,y1-1,cur+1); p[x1][y1]=0; }
}