Camelot USACO 3.3（阵亡）
好烦的一题，思路不是很难，bfs,数据结构要定义好，写起来各种问题，看nocow题解，照某个敲，到case20就是不过。
我擦，后来我直接把他代码复制过去都过不了。找了个别人的先过了...
1 /* 2 3 ID: hubiao cave 4 5 PROG: camelot 6 7 LANG: C++ 8 9 */ 10 11 #include <iostream> 12 #include <fstream> 13 #include<cstring> 14 15 using namespace std; 16 17 struct data{ 18 int x,y,d; 19 }q[10000]; 20 21 bool use[40][40]; 22 int l,r,n,m,x[1001],y[1001],kx,ky,d[40][40][40][40],tt=0,res=2147483647,step[8][2]={{1,2},{2,1},{-1,2},{2,-1},{1,-2},{-2,1},{-1,-2},{-2,-1}}; 23 24 int main() 25 { 26 ifstream fin("camelot.in"); 27 ofstream fout("camelot.out"); 28 29 char cy; 30 fin>>n>>m; 31 fin>>cy>>kx; 32 ky=cy-'A'+1; 33 while (!fin.eof()) 34 { 35 tt++; 36 fin>>cy>>x[tt]; 37 y[tt]=cy-'A'+1; 38 } 39 tt++; 40 for (int i=1;i<=n;i++) 41 for (int j=1;j<=m;j++) 42 for (int k=1;k<=n;k++) 43 for (int t=1;t<=m;t++) 44 d[i][j][k][t]=10000000; 45 for (int i=1;i<=n;i++) 46 for (int j=1;j<=m;j++) 47 { 48 memset(q,0,sizeof(q)); 49 memset(use,false,sizeof(use)); 50 l=1; 51 r=1; 52 q[1].x=i; 53 q[1].y=j; 54 use[i][j]=true; 55 d[i][j][i][j]=0; 56 for(;;) 57 { 58 for (int k=0;k<=7;k++) 59 { 60 int xx=q[l].x+step[k][0],yy=q[l].y+step[k][1]; 61 if (((xx>=1)&&(xx<=n)&&(yy>=1)&&(yy<=m)&&(!use[xx][yy]))){ 62 r++; 63 q[r].x=xx; 64 q[r].y=yy; 65 d[i][j][xx][yy]=q[l].d+1; 66 q[r].d=d[i][j][xx][yy]; 67 use[xx][yy]=true; 68 } 69 } 70 l++; 71 if (l>r) break; 72 } 73 } 74 75 for (int i=1;i<=n;i++) 76 for (int j=1;j<=m;j++) 77 { 78 int minn=2147483647,sum=0; 79 for (int k=1;k<=tt;k++) 80 sum+=d[i][j][x[k]][y[k]]; 81 if (sum+max(abs(kx-i),abs(ky-j))<minn) minn=sum+max(abs(kx-i),abs(ky-j)); 82 for (int xx=max(1,kx-2);xx<=min(n,kx+2);xx++) 83 for (int yy=max(1,ky-2);yy<=min(m,ky+2);yy++) 84 for (int k=1;k<=tt;k++) 85 { 86 int now=sum-d[i][j][x[k]][y[k]]+d[x[k]][y[k]][xx][yy]+d[xx][yy][i][j]+max(abs(kx-xx),abs(ky-yy)); 87 if (now<minn) minn=now; 88 } 89 if (minn<res) res=minn; 90 } 91 fout<<res<<endl; 92 return 0; 93 }
