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紫书 习题3-5

时间:2017-07-19 00:11:28      阅读:200      评论:0      收藏:0      [点我收藏+]

标签:char s   break   ace   ati   name   str   algo   main   cst   

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const char inst[] = "ABLR";
const int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

int main(void)
{
    int t = 0;
    char s[5][6];
    char c;
    while ((s[0][0] = getchar()) != ‘Z‘) {
        int bi = 0, bj = 0;
        for (int i = 0; i < 5; i ++) {
            for (int j = 0; j < 5; j ++) {
                if (!i && !j) continue;
                s[i][j] = getchar();
                if (s[i][j] == ‘ ‘) {bi = i, bj = j;}
            }
            getchar();
        }
        bool legal = true;
        while ((c = getchar()) != ‘0‘) {
            if (legal == false || c == ‘\n‘) continue;
            int k;
            for (k = 0; k < 4; k ++) {
                if (c == inst[k]) break;
            }
            if (k == 4)
                legal = false;
            else {
                int ni = bi+dir[k][0], nj = bj+dir[k][1];
                if (0 <= ni && ni < 5 && 0 <= nj && nj < 5) {
                    swap(s[bi][bj], s[ni][nj]);
                    bi = ni, bj = nj;
                } else
                    legal = false;
            }
        }
        if (++t > 1) printf("\n");
        printf("Puzzle #%d:\n", t);
        if (legal == false)
            printf("This puzzle has no final configuration.\n");
        else {
            for (int i = 0; i < 5; i ++) {
                for (int j = 0; j < 5; j ++) {
                    printf("%c%c", s[i][j], j == 4 ? ‘\n‘ : ‘ ‘);
                }
            }
        }
        getchar();
    }

    return 0;
}

  

紫书 习题3-5

标签:char s   break   ace   ati   name   str   algo   main   cst   

原文地址:http://www.cnblogs.com/Roni-i/p/7203402.html

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