标签：log memset include pac freopen sizeof 二进制 open article
参考blog：http://blog.csdn.net/u013480600/article/details/25336473
考场60RE 后来大家都在讨论怎么判重的时候才想起来我根本没有判重23333
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<algorithm> 6 #include<string> 7 using namespace std; 8 int yx[4]={0,1,-1,0}; 9 int yy[4]={1,0,0,-1}; 10 int n,m,l; 11 bool vis[30][30]; 12 bool c[30][30]; 13 int d[200000]; 14 struct zb{ 15 int l,r; 16 }s[200000][10]; 17 int bfs(){ 18 int h=0,t=1; 19 d[1]=0; 20 while (h<=t){ 21 h++; 22 int x=s[h][1].l,y=s[h][1].r; 23 if (x==1&&y==1) return d[h]; 24 memset(c,0,sizeof(c)); 25 for (int i=1;i<=l;++i) c[s[h][i].l][s[h][i].r]=1; 26 for (int i=0;i<=3;++i){ 27 if (x+yx[i]<=n&&x+yx[i]>=1&&y+yy[i]<=m&&y+yy[i]>=1&&(!vis[x+yx[i]][y+yy[i]])&&(!c[x+yx[i]][y+yy[i]])){ 28 t++; 29 d[t]=d[h]+1; 30 s[t][1].l=x+yx[i],s[t][1].r=y+yy[i]; 31 for (int i=2;i<=l;++i) s[t][i].l=s[h][i-1].l,s[t][i].r=s[h][i-1].r; 32 } 33 } 34 } 35 36 } 37 int main(){ 38 freopen ("snake.in","r",stdin); 39 freopen ("snake.out","w",stdout); 40 scanf ("%d%d%d",&n,&m,&l); 41 for (int i=1;i<=l;++i) scanf ("%d%d",&s[1][i].l,&s[1][i].r); 42 int k; 43 scanf ("%d",&k); 44 int x,y; 45 for (int i=1;i<=k;++i) scanf ("%d%d",&x,&y),vis[x][y]=1; 46 printf("%d",bfs()); 47 return 0; 48 }
然后改了好久啊啊啊啊啊啊哭泣
跑出来结果还很难看 不过至少是过了
这题的精髓在于将蛇身各段之间的关系以二进制的形式表现出来 然后就是利用二进制进行状态判重以及对是否会与自己发生冲突的判断
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 #include<algorithm> 6 #include<string> 7 using namespace std; 8 int yx[4]={1,0,-1,0}; 9 int yy[4]={0,-1,0,1}; 10 int n,m,l; 11 bool viss[30][30][1<<15]; 12 bool vis[30][30]; 13 int lin[20]; 14 struct zb{ 15 int l,r,z,d; 16 }s[800000]; 17 bool check(int x,int y,int s,int p,int q){ 18 int i=l-1; 19 for (int i=l-1;i>=1;--i) lin[i]=s&3,s>>=2; 20 for (int i=1;i<l;++i){ 21 if (p+yx[lin[i]]==x&&q+yy[lin[i]]==y) return 0; 22 p+=yx[lin[i]],q+=yy[lin[i]]; 23 } 24 return 1; 25 } 26 int bfs(){ 27 int h=0,t=1; 28 while (h<=t){ 29 h++; 30 int x=s[h].l,y=s[h].r,si=s[h].z; 31 viss[x][y][si]=1; 32 if (x==1&&y==1) return s[h].d; 33 for (int i=0;i<=3;++i){ 34 if (x+yx[i]<=n&&x+yx[i]>=1&&y+yy[i]<=m&&y+yy[i]>=1&&(!vis[x+yx[i]][y+yy[i]])&&check(x+yx[i],y+yy[i],si,x,y)){ 35 int st=(si>>2)+(((i+2)%4)<<((l-2)*2)); 36 if (viss[x+yx[i]][y+yy[i]][st]) continue; 37 if (x+yx[i]==1&&y+yy[i]==1) return s[h].d+1; 38 //cout<<x<<" "<<y<<" "<<si<<" "<<" "<<st<<" "<<i<<endl; 39 viss[x+yx[i]][y+yy[i]][st]=1; 40 t++; 41 s[t].d=s[h].d+1,s[t].z=st; 42 s[t].l=x+yx[i],s[t].r=y+yy[i]; 43 } 44 } 45 } 46 47 } 48 int main(){ 49 freopen ("snake.in","r",stdin); 50 freopen ("snak.out","w",stdout); 51 scanf ("%d%d%d",&n,&m,&l); 52 scanf ("%d%d",&s[1].l,&s[1].r); 53 int xx=s[1].l,yyy=s[1].r; 54 for (int i=2;i<=l;++i){ 55 int x,y; 56 scanf ("%d%d",&x,&y); 57 for (int j=0;j<=3;++j){ 58 if (xx+yx[j]==x&&yyy+yy[j]==y){ 59 s[1].z=(s[1].z<<2)+j; 60 break; 61 } 62 } 63 xx=x,yyy=y; 64 } 65 s[1].d=0; 66 int k; 67 scanf ("%d",&k); 68 int x,y; 69 for (int i=1;i<=k;++i) scanf ("%d%d",&x,&y),vis[x][y]=1; 70 printf("%d",bfs()); 71 return 0; 72 }
大家好像都跟着博客用的队列可是我实在是懒飞 结果慢了好多好多而且空间也大了好多好多
不过对于我这种辣鸡过了就很开心啦诶嘿嘿嘿嘿
标签：log memset include pac freopen sizeof 二进制 open article
原文地址：http://www.cnblogs.com/YZSTUDIO/p/6536021.html
