UVa 101 The Blocks Problem

【解题思路】

本题题意比较明确，只要了解四个指令的含义即可（以下a和b都是方块的编号），可分成四个函数编写。

move a onto b：将方块a移到方块b相邻的上方，在此之前，a和b之上的方块都要按照编号移回到原来的位置。

move a over b：将方块a移到方块b所在的堆上，在此之前，a之上的方块都要按照编号移回到原来的位置。

pile a onto b：将方块a以及之上的方块顺序不变地移到方块b相邻的上方，在此之前，b之上的方块都要按照编号移回到原来的位置。

pile a over b：将方块a以及之上的方块顺序不变地移到方块b所在的堆上。

最后题目输出部分每一行的开头是没有空格的，差点被样例输出坑到。

【代码】

#include<stdio.h>#include<string.h>typedef struct tagStack{int num[30];//记录该堆的方块编号int len;//该堆中方块的数量int loc;//指示该堆的编号对应的方块在哪一个堆}Stack;Stack stack[30];int searchLoc(int syn) { //在堆中搜索方块syn在哪一层，返回位置int p;int loc = stack[syn].loc;for (p = 0; p < stack[loc].len; p ++) {if (stack[loc].num[p] == syn)break;}return p;}void MoveOnto(int a, int b) { //move a onto b//四个函数的功能大同小异，这里只解释一个函数里的功能if (stack[a].loc != stack[b].loc) { //a和b两个方块所在堆不同时，指令有效int pa, pb, loca, locb;pa = searchLoc(a);pb = searchLoc(b);loca = stack[a].loc;locb = stack[b].loc;for (int i = stack[locb].len-1; i > pb; i --) { //移走方块b上方的所有方块，移回原来位置int t = stack[locb].num[i];stack[t].num[stack[t].len ++] = t;stack[t].loc = t;}stack[locb].len = pb+1;//修改方块b所在堆的方块个数for (int i = stack[loca].len-1; i > pa; i --) { //移走方块a上方的所有方块，移回原来位置int t = stack[loca].num[i];stack[t].num[stack[t].len ++] = t;stack[t].loc = t;}stack[loca].len = pa;//修改方块a所在堆的方块个数stack[locb].num[stack[locb].len ++] = stack[loca].num[pa]; //将方块a移动方块b的顶部，b所在堆个数加1stack[a].loc = locb;//方块a所在堆修改为方块b的所在堆}}void MoveOver(int a, int b) { //move a over bif (stack[a].loc != stack[b].loc) {int pa, pb, loca, locb;pa = searchLoc(a);pb = searchLoc(b);loca = stack[a].loc;locb = stack[b].loc;for (int i = stack[loca].len-1; i > pa; i --) {int t = stack[loca].num[i];stack[t].num[stack[t].len ++] = t;stack[t].loc = t;}stack[loca].len = pa;stack[locb].num[stack[locb].len ++] = stack[loca].num[pa];stack[a].loc = locb;}}void PileOnto(int a, int b) { //pile a onto bif (stack[a].loc != stack[b].loc) {int pa, pb, loca, locb;pa = searchLoc(a);pb = searchLoc(b);loca = stack[a].loc;locb = stack[b].loc;for (int i = stack[locb].len-1; i > pb; i --) {int t = stack[locb].num[i];stack[t].num[stack[t].len ++] = t;stack[t].loc = t;}stack[locb].len = pb+1;for (int i = pa; i < stack[loca].len; i ++) {int t = stack[loca].num[i];stack[locb].num[stack[locb].len ++] = t;stack[t].loc = locb;}stack[loca].len = pa;}}void PileOver(int a, int b) { //pile a over bif (stack[a].loc != stack[b].loc) {int pa, pb, loca, locb;pa = searchLoc(a);pb = searchLoc(b);loca = stack[a].loc;locb = stack[b].loc;for (int i = pa; i < stack[loca].len; i ++) {int t = stack[loca].num[i];stack[locb].num[stack[locb].len ++] = t;stack[t].loc = locb;}stack[loca].len = pa;}}int main() {int n, a, b;char cmd1[10], cmd2[10];while (~scanf("%d", &n)) {for (int i = 0; i < n; i ++) {stack[i].num[0] = i;stack[i].len = 1;stack[i].loc = i;}while (true) {scanf("%s", cmd1);if (strcmp(cmd1, "quit") == 0) {break;

<span style="white-space:pre"></span>}scanf("%d %s %d", &a, cmd2, &b);if (strcmp(cmd1, "move") == 0) {if (strcmp(cmd2, "onto") == 0)MoveOnto(a, b);elseMoveOver(a, b);}else {if (strcmp(cmd2, "onto") == 0)PileOnto(a, b);elsePileOver(a, b);}}for (int i = 0 ; i < n; i ++) {printf("%d:", i);for (int j = 0; j < stack[i].len; j ++)printf(" %d", stack[i].num[j]);printf("\n");}}return 0;}