IOI 98
Centuries ago, King Arthur and the Knights of the Round Table used to meet every year on New Year's Day to celebrate their fellowship. In remembrance of these events, we consider a board game for one player, on which one chesspiece king and several knight pieces are placed on squares, no two knights on the same square.
This example board is the standard 8x8 array of squares:
The King can move to any adjacent square from 
to 
as long as it does not fall off the board:
A Knight can jump from to , as long as it does not fall off the board:
During the play, the player can place more than one piece in the same square. The board squares are assumed big enough so that a piece is never an obstacle for any other piece to move freely.
The player's goal is to move the pieces so as to gather them all in the same square - in the minimal number of moves. To achieve this, he must move the pieces as prescribed above. Additionally, whenever the king and one or more knights are placed in the same square, the player may choose to move the king and one of the knights together from that point on, as a single knight, up to the final gathering point. Moving the knight together with the king counts as a single move.
Write a program to compute the minimum number of moves the player must perform to produce the gathering. The pieces can gather on any square, of course.
PROGRAM NAME: camelot
INPUT FORMAT
| Line 1: | Two space-separated integers: R,C, the number of rows and columns on the board. There will be no more than 26 columns and no more than 30 rows. |
| Line 2..end: | The input file contains a sequence of space-separated letter/digit pairs, 1 or more per line. The first pair represents the board position of the king; subsequent pairs represent positions of knights. There might be 0 knights or the knights might fill the board. Rows are numbered starting at 1; columns are specified as upper case characters starting with `A'. |
SAMPLE INPUT (file camelot.in)
8 8
D 4
A 3 A 8
H 1 H 8
The king is positioned at D4. There are four knights, positioned at A3, A8, H1, and H8.
OUTPUT FORMAT
A single line with the number of moves to aggregate the pieces.
SAMPLE OUTPUT (file camelot.out)
10
SAMPLE OUTPUT ELABORATION
They gather at B5.
Knight 1: A3 - B5 (1 move)
Knight 2: A8 - C7 - B5 (2 moves)
Knight 3: H1 - G3 - F5 - D4 (picking up king) - B5 (4 moves)
Knight 4: H8 - F7 - D6 - B5 (3 moves)
1 + 2 + 4 + 3 = 10 moves.
题解：BFS+最优剪枝。自己YY了个想法，就是先用BFS（Floyd会超时。。。）算出所有点对的最短距离，接着枚举集结点，然后计算出所有骑士到集结点的距离和，之后枚举每一个骑士带上国王一起走的方案（可能需要国王走一定的距离），就是这步把我给搞蛋疼了。。。因为我的想法是先让国王走一段距离，然后让骑士带走，这样的话时间复杂度非常高。。。明显又得超时。。。。无奈只好去看解题报告。。。。网上的方法说接国王的点为国王所在位置偏移两个距离就行（即+-2），但貌似无法证明它的正确性，可能是数据不够强大的原因吧。。。所以也能AC。不过这样的做法也可能有它的合理性，让国王尽量少走路，因为骑士每步的偏移量大些，所以尽量让骑士走。提交了几次，第一次是输入有问题，第二次是数组越界，第三次是因为把数组d清成0了，这样也会WA.
例如这组数据:
2 26
A 1
Y 1
有些集结点骑士根本不能达到，但是还是会进行处理，它会把骑士走的距离计算成0，这样就造成错误了。。。所以必须把d数组全部初始化为一个大值才行。这个错误说明初始化也是非常的重要，一定要看清楚题，再进行相应的初始化，如果不注意的话，到时比赛很有可能就悲剧了。。。
View Code
1 /* 2 ID:spcjv51 3 PROG:camelot 4 LANG:C 5 */ 6 #include<stdio.h> 7 #include<string.h> 8 #include<stdlib.h> 9 #include<math.h> 10 #define MAXX 35 11 #define MAXY 35 12 #define MAXS 10000 13 int dx[8]= {2,2,1,1,-2,-2,-1,-1}; 14 int dy[8]= {-1,1,-2,2,-1,1,-2,2}; 15 int g[MAXX][MAXY],visit[MAXX][MAXY]; 16 long d[MAXX][MAXY][MAXX][MAXY]; 17 int m,n,kx,ky; 18 long mins; 19 long max(long a,long b) 20 { 21 return a>b?a:b; 22 } 23 long min(long a,long b) 24 { 25 return a<b?a:b; 26 } 27 void shortpath(int sx,int sy) 28 { 29 long q[2][1000]; 30 long head,tail,i,tx,ty; 31 memset(q,0,sizeof(q)); 32 memset(visit,0,sizeof(visit)); 33 head=0; 34 tail=1; 35 q[0][tail]=sx; 36 q[1][tail]=sy; 37 visit[sx][sy]=1; 38 d[sx][sy][sx][sy]=0; 39 while(head<tail) 40 { 41 head++; 42 tx=q[0][head]; 43 ty=q[1][head]; 44 for(i=0; i<8; i++) 45 if(tx+dx[i]>=1&&tx+dx[i]<=n&&ty+dy[i]>=1&&ty+dy[i]<=m&&!visit[tx+dx[i]][ty+dy[i]]) 46 { 47 tail++; 48 q[0][tail]=tx+dx[i]; 49 q[1][tail]=ty+dy[i]; 50 visit[tx+dx[i]][ty+dy[i]]=1; 51 d[sx][sy][q[0][tail]][q[1][tail]]=d[sx][sy][tx][ty]+1; 52 } 53 } 54 } 55 long solve(int x,int y) 56 { 57 long i,j,sum,tt,tx,ty; 58 sum=0; 59 for(i=1;i<=n;i++) 60 for(j=1;j<=m;j++) 61 if(g[i][j]) 62 { 63 sum+=d[i][j][x][y]; 64 if(sum>mins) return MAXS; 65 } 66 tt=sum+max(abs(x-kx),abs(y-ky)); 67 for(tx=max(1,kx-2);tx<=min(n,kx+2);tx++) 68 for(ty=max(1,ky-2);ty<=min(m,ky+2);ty++) 69 for(i=1;i<=n;i++) 70 for(j=1;j<=m;j++) 71 if(g[i][j]) 72 tt=min(tt,sum+max(abs(tx-kx),abs(ty-ky))+d[i][j][tx][ty]+d[tx][ty][x][y]-d[i][j][x][y]); 73 //printf("%d\n",tt); 74 return tt; 75 } 76 int main(void) 77 { 78 freopen("camelot.in","r",stdin); 79 freopen("camelot.out","w",stdout); 80 int i,j,t; 81 long ans,ts; 82 char ch[10]; 83 scanf("%d%d",&n,&m); 84 scanf("%s%d",ch,&t); 85 kx=t; 86 ky=ch[0]-'A'+1; 87 memset(g,0,sizeof(g)); 88 memset(d,1,sizeof(d)); 89 while(scanf("%s%d",ch,&t)!=EOF) 90 { 91 g[t][ch[0]-'A'+1]=1; 92 getchar(); 93 } 94 for(i=1; i<=n; i++) 95 for(j=1; j<=m; j++) 96 shortpath(i,j); 97 ans=MAXS; 98 mins=MAXS; 99 for(i=1; i<=n; i++) 100 for(j=1; j<=m; j++) 101 { 102 ts=solve(i,j); 103 if(ts<ans) 104 { 105 ans=ts; 106 mins=ans; 107 } 108 } 109 printf("%ld\n",ans); 110 return 0; 111 }