【笔记】《算法竞赛入门》习题7-3 UVa211_多米诺效应
题目: 使用28个多米诺骨牌拼凑出输入的二维数组
思路,使用一个二维数组标记每个多米诺骨牌的牌号,下标为Pips,一个典型的回溯
//
// Created by hsby on 2021/1/30.
//
#include <iostream>
#include <cstring>
using namespace std;
int Dir[2][2] = {{0, 1}, {1, 0}};
int dom[7][7] = {0};
int g[7][8];
int ans[7][8];
int t;
bool judge(){
for (int i = 0; i < 7; ++i) {
for (int j = 0; j < 8; ++j) {
if (!ans[i][j]) return false;
}
}
return true;
}
void dfs(int ai){
// 如果判断可行,输出结果
if (judge()){
for (int i = 0; i < 7; ++i) {
for (int j = 0; j < 8; ++j) {
printf("%2d ", ans[i][j]);
}
cout << endl;
}
cout << endl;
t++;
return;
}
// flag表示在ai行之后找到了可以继续遍历的起点
int flag = 1;
// 从ai行开始找
for (int i = ai; i < 7 && flag; ++i) {
for (int j = 0; j < 8 && flag; ++j) {
if (ans[i][j]) continue;
// 找到了,开始往右和下两个方向探索
for (int k = 0; k < 2; ++k) {
int ti = i + Dir[k][0];
int tj = j + Dir[k][1];
if (ans[ti][tj] || ti > 6 || ti < 0 || tj > 7 || tj < 0) continue;
flag = 0;
int a = min(g[i][j], g[ti][tj]);
int b = max(g[i][j], g[ti][tj]);
int d = dom[a][b];
ans[i][j] = d;
ans[ti][tj] = d;
dfs(i);
ans[i][j] = 0;
ans[ti][tj] = 0;
}
}
}
}
int main(){
int cnt = 1;
for (int i = 0; i < 7; ++i) {
for (int j = i; j < 7; ++j) {
dom[i][j] = cnt++;
}
}
int n = 1;
while (!cin.eof()){
memset(g, 0, sizeof(g));
memset(ans, 0, sizeof(ans));
t = 0;
cout << "Layout #" << n << ":" << endl << endl;
for (int i = 0; i < 7; ++i) {
for (int j = 0; j < 8; ++j) {
cin >> g[i][j];
cout << g[i][j] << " ";
}
cout << endl;
}
cout << endl;
cout << "Maps resulting from layout #" << n << " are:" << endl << endl;
dfs(0);
cout << "There are " << t++ << " solution(s) for layout #" << n++ << "." << endl << endl;
}
return 0;
}