您的位置:首页 > 其它

Sicily 1152. 简单的马周游问题

2014-10-10 09:43 337 查看
题目链接在此

这个题解未采用类似“启发式搜索”的做法。只是po一下代码。

更快捷的做法与详解见另一篇博文:《马的周游问题》


#include<iostream>
#include<vector>

using namespace std;

struct pos {
int row;
int col;

pos() {}
pos(int r, int c) : row(r), col(c) {}

pos plus(pos a) {
return pos(a.row + this->row, a.col + this->col);
}
};

bool visited[31];
int printSeq[31];
const struct pos available[] = { pos(1, -2), pos(2, -1), pos(2, 1), pos(1, 2),
pos(-1, 2), pos(-2, 1), pos(-2, -1), pos(-1, -2) };

struct pos one_to_two_dim(int n) {
int r = (n - 1) / 6 + 1;
int c = (n - 1) % 6 + 1;
return pos(r, c);
}

int two_to_one_dim(struct pos p) {
if (p.row >= 1 && p.row <= 5 && p.col >= 1 && p.col <= 6)
return (p.row - 1) * 6 + p.col;
else
return 0;
}

void DFS(int n, bool& isDone, int counter) {
if (isDone)
return;

if (counter == 31) {
isDone = true;

bool spaceFlag = false;
for (int i = 1; i <= 30; i++) {
if (!spaceFlag)
spaceFlag = true;
else
cout << ' ';

cout << printSeq[i];
}
cout << endl;
}
else {
for (int i = 0; i < 8; i++) {
struct pos tmp = one_to_two_dim(n);
int going_to = two_to_one_dim(tmp.plus(available[i]));

if (going_to >= 1 && going_to <= 30 && visited[going_to] == false) {
visited[going_to] = true;
printSeq[counter] = going_to;
counter++;

DFS(going_to, isDone, counter);

visited[going_to] = false;
counter--;
}
}
}
}

int main() {
int start;
cin >> start;

while (start != -1) {
bool isDone = false;

for (int i = 1; i <= 30; i++) {
printSeq[i] = 0;
visited[i] = false;
}

printSeq[1] = start;
visited[start] = true;

DFS(start, isDone, 2);

cin >> start;
}

return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: