人工智能及其应用 实验课 源代码

Posted 2023-03-08 行码棋

实验一

梵塔问题

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
ll ans = 0;
void dfs(char a, char b, char c, int n)

	if(n == 1)
	
		ans++;
		cout << a << " -> " << c << "\\n";
	
	else
	
		dfs(a, c, b, n - 1);
		ans++;
		cout << a << " -> " << c << "\\n";
		dfs(b, a, c, n - 1);
	

int main()

	ios::sync_with_stdio(false);
	cin.tie(0);
	int n;
	cin >> n;
	dfs('A', 'B', 'C', n);
	cout << ans << "\\n";
	return 0;

传教士野人渡河问题

#include<bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
const int dx[] = 1, 0, 0, 2, 1;
const int dy[] = 1, 2, 1, 0, 0;

array<int, 3> ans[N], tot[N];
map<array<int, 3>, bool> v;
int sum, tt = -1;
int a, b;

// 本岸位置的 野人 和 传教士 数量  1 ： 本岸   -1 ：对岸 
bool check(array<int, 3> t)

	int y = t[0], c = t[1], f = t[2];
	if(y < 0 || c < 0 || y > a || c > b) return 0;
	if((c != 0 && y > c) || (b - c != 0 && a - y > b - c)) return 0;
	return 1;

void dfs(array<int, 3> t)

	int y = t[0], c = t[1], f = t[2];
	if(y == 0 && c == 0 && f == -1)
	
		sum++;
		cout << sum << "\\n";
		for(int i = 0; i <= tt; i++)
		
			cout << (ans[i][2] == 1 ? "去" : "回") << " ";
			cout << "野人:" << ans[i][0] << " 传教士:" << ans[i][1] << "\\n";
		
		return;
	
	array<int, 3> tmp;
	for(int i = 0; i < 5; i++)
	
		tmp = y - f * dx[i], c - f * dy[i], -f;
		if(check(tmp) && !v[tmp])
		
			v[tmp] = 1;
			ans[++tt] = dx[i], dy[i], f;
			dfs(tmp);
			v[tmp] = 0;
			--tt;
		
	

int main()

	cout << "请分别输入野人数量和传教士数量:";
	cin >> a >> b;
	array<int, 3> t = a, b, 1;
	v[t] = 1;
	dfs(t);
	v[t] = 0;
	return 0;

实验二

宽度优先和深度优先求解路径

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, M = 2 * N;
const int inf = 0x3f3f3f3f;

int n, m;
int s, d;
int dis[N], vis[N], pre[N];

int e[M], h[N], ne[M], w[M], idx;
void add(int a, int b, int c)

	e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;

void dfs(int u)

	vis[u] = 1;
	if(u == d) return;
	for(int i = h[u]; ~i; i = ne[i])
	
		int to = e[i], val = w[i];
		if(!vis[to] || dis[u] + val < dis[to])
		
			dis[to] = dis[u] + val;
			pre[to] = u;
			dfs(to);
		
	

// 只能求不带权图 
void bfs(int s)

	queue<int> q;
	vis[s] = 1;
	q.push(s);
	while(!q.empty())
	
		int t = q.front();
		q.pop();
		if(t == d) return;
		for(int i = h[t]; ~i; i = ne[i])
		
			int to = e[i];
			if(vis[to]) continue;
			vis[to] = 1;
			dis[to] = dis[t] + 1;
			pre[to] = t;
			q.push(to);
		
	

int main()

	ios::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	
	memset(h, -1, sizeof h);
	cin >> n >> m >> s >> d;
	for(int i = 1; i <= m; i++)
	
		int a, b, c;
		cin >> a >> b >> c;
		add(a, b, c);
	
	
	memset(dis, 0x3f, sizeof dis);
	memset(pre, -1, sizeof pre);
	dis[s] = 0;
//	dfs(s);
	bfs(s);
	vector<int> path;
	int t = d;
	path.push_back(t);
	while(pre[t] != -1)
	
		path.push_back(pre[t]);
		t = pre[t];
	
	reverse(path.begin(), path.end());
	cout << "最短的路径:\\n";
	for(int i = 0; i < int(path.size() - 1); i++)
		cout << path[i] << " -> " << path[i + 1] << "\\n";
	cout << "最短路径:" << dis[d] << "\\n";
	return 0;


/* sample 1 
8 10 1 8
1 2 1
1 5 1
2 5 1
2 3 1
5 3 1
5 6 1
3 4 1
3 7 1
3 8 1
4 8 1
*/

八数码问题

下面代码实现了N数码问题，不仅仅局限于八数码，但是求解效率有限

#include<bits/stdc++.h>
using namespace std;
using pis = pair<int, string>;
const int N = 500 * 500 + 5;
const int dx[] = -1, 0, 1, 0, dy[] = 0, 1, 0, -1;
int n;

template <typename T>
struct Fenwick

	vector<T> tr;
	const int n;
	Fenwick(int n): n(n), tr(n) 
	void insert(int x, T d)
	
		for(; x < n; x += x & (-x))
			tr[x] += d;
	
	T sum(int x)
	
		T res = 0;
		for(; x; x -= x & (-x))
			res += tr[x];
		return res;
	
	T RangeSum(int l, int r)
	
		return sum(r) - sum(l - 1);
	
;

int f(string s)

	int cnt = 0;
	for(int i = 0; i < int(s.size()); i++)
		if(s[i] != '0')
		
			int cur = s[i] - '1';
			cnt += abs(cur / n - i / n) + abs(cur % n - i % n);
		
	return cnt;

void bfs(string start, string end)

	char op[] = "URDL";
	
	unordered_map<string, int> dis;
	unordered_map<string, pair<char, string>> pre;
	priority_queue<pis, vector<pis>, greater<pis>> q;
	
	q.push(f(start), start);
	dis[start] = 0;
	while(!q.empty())
	
		auto t = q.top();
		q.pop();
		string st = t.second;
		
		if(st == end) break;
		
		int x, y;
		for(int i = 0; i < int(st.size()); i++)
			if(st[i] == '0')
			
				x = i / n, y = i % n;
				break;
			
		
		for(int i = 0; i < 4; i++)
		
			int nx = x + dx[i], ny = y + dy[i];
			if(nx < 0 || nx >= n || ny < 0 || ny >= n) continue;
			string t = st;
			swap(t[x * n + y], t[nx * n + ny]);
			if(!dis.count(t) || dis[st] + 1 < dis[t])
			
				dis[t] = dis[st] + 1;
				pre[t] = op[i], st;
				q.push(dis[t] + f(t), t);
			
		
	
	string move;
	vector<string> state;
	state.push_back(end);
	while(end != start)
	
		move += pre[end].first;
		end = pre[end].second;
		state.push_back(end);
	
	reverse(move.begin(), move.end());
	reverse(state.begin(), state.end());
	
	for(int i = 0; i < int(move.size()); i++)
	
		int cur = 0;
		for(int j = 0; j < n; j++)
			for(int k = 0; k < n; k++)
				cout << (state[i][cur++] - '0') << " \\n"[k == n - 1];
		
		cout << "-----------------\\n操作:" << move[i] << "\\n";
		if(i == int(move.size() - 1))
		
			int nxt = i + 1;
			cur = 0;
			for(int j = 0; j < n; j++)
				for(int k = 0; k < n; k++)
					cout << (state[nxt][cur++] - '0') << " \\n"[k == n - 1];
		
	
	

int main()

	cout << "输入数码阶数:\\n"; 
	cin >> n;
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