2016-08-18

UVA - 101 The Blocks Problem

题目大意：操作积木。有 0~n-1 个积木，初始状态是并排横放。有四种操作：move a onto b，将 a、b 上的方块放回原位然后将 a 放在 b 上；move a over b，将 a 上的方块放回原味然后将 a 放在 b 所在位置的最顶上；pile a onto b，将 b 上的方块放回原味然后将 a 及其上所有的方块放到 b 上；pile a over b，将 a 及其上所有方块放到 b 所在的位置最顶上。

解题思路：开个二维数组模拟就好，这种题目把最小的动作分解出来组合就好了。

注意：这题主要是要注意一些小细节，一直WA。。。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int map[50][50];char str[2][20];int a, b;void find(int x) {for (int i = 0; i < 25; i++)for (int j = 0; j < 25; j++)if ( map[i][j] == x ) {a = i;b = j;}}void del(int x) {find(x);for (int i = b; i < 25; i++)        map[a][i] = -1;}void re(int x) {find(x);for (int i = b+1; i < 25; i++)if ( map[a][i] != -1 ) {int tmp = map[a][i];map[tmp][0] = tmp;map[a][i] = -1;}}void move_onto(int x, int y) {re(x);re(y);find(x);map[a][b] = -1;find(y);map[a][b+1] = x;}void move_over(int x, int y) {re(x);del(x);find(y);for (int i = b; i < 25; i++)if ( map[a][i] == -1) {map[a][i] = x;break;}}void pile_onto(int x, int y) {re(y);find(x);int tmp[50];int j = 0;for (int i = b; map[a][i] != -1; i++)tmp[j++] = map[a][i];int tag = j;del(x);find(y);j = 0;for (int i = b; j < tag; i++)if ( map[a][i] == -1 )map[a][i] = tmp[j++];}void pile_over(int x, int y) {find(x);int tmp[50];int j = 0;for (int i = b; map[a][i] != -1; i++)tmp[j++] = map[a][i];int tag = j;del(x);find(y);j = 0;for (int i = b; j < tag; i++)if ( map[a][i] == -1 ) map[a][i] = tmp[j++];}int main() {int n;int c, d;scanf("%d", &n);memset(map, -1, sizeof(map));for (int i = 0; i < n; i++)map[i][0] = i;while (scanf("%s", str[0]) ) {if ( !strcmp(str[0], "quit") )break;scanf("%d%s%d", &c, str[1], &d);find(c);int x1 = a;find(d);int x2 = a;if ( x1 == x2 )continue;if ( !strcmp(str[0], "move") && !strcmp(str[1], "onto") )move_onto(c, d);if ( !strcmp(str[0], "move") && !strcmp(str[1], "over") )move_over(c, d);if ( !strcmp(str[0], "pile") && !strcmp(str[1], "onto") )pile_onto(c, d);if ( !strcmp(str[0], "pile") && !strcmp(str[1], "over") )pile_over(c, d);}for (int i = 0; i < n; i++) {cout << i << ":";for (int j = 0; map[i][j] != -1; j++)cout << " " << map[i][j];cout << endl;}return 0;}