思路
- 這一題我們用bfs來搜索每一條路徑,對與正在搜素的某個路徑的下一個節點位置我們怎麼判斷它是否已經走過了?,這做的思路就是對每個節點 我們都給 加一個 vector road; 去存儲路徑,如果路徑road中已經有這個節點那麼表 這個節點已經走過了,那麼這條路徑就重複了,就不要把這個路徑壓入到 隊列q 中了,,,,還有一下剪枝、細節直接看代碼吧
代碼
#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>
#include<map>
#include<string>
#include<cstdio>
#include<cmath>
#include<stack>
void fre() { freopen("A.txt", "r", stdin); freopen("Ans.txt","w",stdout); } void Fre() { freopen("A.txt", "r", stdin);}
#define ios ios::sync_with_stdio(false)
#define Pi acos(-1)
#define pb push_back
#define fi first
#define se second
#define ll long long
#define ull unsigned long long
#define db double
#define Pir pair<int, int>
#define PIR pair<Pir, Pir>
#define m_p make_pair
#define INF 0x3f3f3f3f
#define mod (ll)(1e9 + 7)
#define for_(i, s, e) for(int i = (s); i <= (e); i ++)
#define rep_(i, e, s) for(int i = (e); i >= (s); i --)
#define sd(a) scanf("%d", &a)
#define sc(a) scanf("%c", &a)
using namespace std;
const int mxn = 1e6;
int mz[25][5];
struct Node
{
int p;
vector<int> road;
};
void bfs(int s)
{
Node u((Node){ s });
queue<Node> q;
q.push(u);
int Case = 1;
while(! q.empty())
{
u = q.front(); q.pop();
if(u.road.size() == 20 && u.road[u.road.size() - 1] == s)
{
printf("%d: %d", Case ++, s);
for(auto x : u.road)
printf(" %d", x);
printf("\n");
}
if(u.road.size() && u.road[u.road.size() - 1] == s) continue;
for_(i, 1, 3)
{
if(find(u.road.begin(), u.road.end(), mz[u.p][i]) != u.road.end()) continue;
Node v((Node){ mz[u.p][i], u.road });
v.road.pb(mz[u.p][i]);
q.push(v);
}
}
}
int main()
{
for_(i, 1, 20)
scanf("%d %d %d", &mz[i][1], &mz[i][2], &mz[i][3]);
int s;
while(sd(s) && s)
bfs(s);
return 0;
}