POJ 1178 Camelot
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 1264 | Accepted: 588 |
Description
The Board is an 8x8 array of squares. The King can move to any adjacent square, as shown in Figure 2, as long as it does not fall off the board. A Knight can jump as shown in Figure 3, 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 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 smallest possible 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 henceforth, 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.
Input
0 <= number of knights <= 63
Output
Sample Input
D4A3A8H1H8
Sample Output
10
Source
/* http://acm.pku.edu.cn/JudgeOnline/problem?id=1178 枚举 + floyd */ #include <iostream> #include <string> #include <cmath> #define MAX_N 64 #define MAX_VAL 100000 #define maxv(a, b) ((a) >= (b) ? (a) : (b)) #define minv(a, b) ((a) <= (b) ? (a) : (b)) using namespace std; int dir[8][2] = {{-1, -2}, {-2, -1}, {-2, 1}, {-1, 2}, {1, -2}, {2, -1}, {2, 1}, {1, 2}}; int distK[2][MAX_N + 1][MAX_N + 1]; int pos[MAX_N + 1][2]; int num = 0; bool inRange(int row, int col) { return row >= 0 && row < 8 && col >= 0 && col < 8; } //初始化格子之间的最短距离, 同时将骑士可走的格子之间的距离设置为1 void init() { int i, j, k, nr, nc, p1, p2; for(i = 0;i < MAX_N; i++) for(j = 0; j < MAX_N; j++) { if(i == j) distK[0][i][j] = distK[1][i][j] = 0; else distK[0][i][j] = distK[1][i][j] = MAX_VAL; } for(i = 0; i < 8; i++) { for(j = 0; j < 8; j++) { for(k = 0; k < 8; k++) { nr = i + dir[k][0]; nc = j + dir[k][1]; if(inRange(nr, nc)) { p1 = i * 8 + j; p2 = nr * 8 + nc; distK[0][p1][p2] = 1; } } } } } //计算任意两个格子之间的最短距离 void floyd() { int i, j, k, dij, dik, dkj, last = 0, cur; for(k = 0; k < MAX_N; k++) { cur = 1 - last; for(i = 0; i < MAX_N; i++) { for(j = 0; j < MAX_N; j++) { if(i == j) continue; dij = distK[last][i][j]; dik = distK[last][i][k]; dkj = distK[last][k][j]; if(dij < MAX_VAL && dij <= dik + dkj) distK[cur][i][j] = dij; else if(dik + dkj < MAX_VAL && dik + dkj < dij) distK[cur][i][j] = dik + dkj; else distK[cur][i][j] = MAX_VAL; } } last = cur; } } void getRes() { int res = MAX_VAL; int i, j, k, distS, d1, d2, row, col; //枚举最终的交汇点 for(i = 0; i < MAX_N; i++) { distS = 0; //计算所有骑士到这个点的距离总和 for(j = 1; j < num; j++) distS += distK[0][i][pos[j][0] * 8 + pos[j][1]]; //枚举所有国王可能和骑士的相遇点 for(j = 0; j < MAX_N; j++) { row = j / 8; col = j - row * 8; //d1是国王到j位置的距离 d1 = maxv(abs(pos[0][0] - row), abs(pos[0][1] - col)); //枚举所有骑士,寻找 这个骑士到j的距离再加上从j到i的距离然后减去骑士到i距离的最小值 d2 = MAX_VAL; for(k = 1; k < num; k++) { int curd = 0; //p为骑士的位置 int p = pos[k][0] * 8 + pos[k][1]; curd = distK[0][p][j] + distK[0][j][i] - distK[0][p][i]; if(curd < d2) d2 = curd; } //distS + d1 + d2是当前总的距离 res = minv(res, distS + d1 + d2); } } cout<<res<<endl; } int main() { int i; string input; cin>>input; num = input.length() / 2; for(i = 0; i < num; i++) { pos[i][1] = input[i * 2] - 'A'; pos[i][0] = input[i * 2 + 1] - '1'; } init(); floyd(); getRes(); return 0; }