蒙特卡洛算法C++代码

以下是一个简单的蒙特卡洛算法的C++代码示例，用于计算圆周率的近似值：

#include <iostream>
#include <cstdlib>
#include <ctime>

using namespace std;

int main()
{
    srand(time(0)); // 初始化随机数种子

    int n = 1000000; // 模拟次数
    int inside = 0; // 圆内点数

    for (int i = 0; i < n; i++)
    {
        double x = (double)rand() / RAND_MAX; // 生成[0,1]之间的随机数
        double y = (double)rand() / RAND_MAX;

        if (x * x + y * y <= 1) // 判断是否在圆内
        {
            inside++;
        }
    }

    double pi = (double)inside / n * 4; // 计算圆周率的近似值
    cout << "Pi is approximately " << pi << endl;

    return 0;
}

该代码通过生成随机的点，判断这些点是否在圆内，从而计算出圆周率的近似值。

相关问题

亚马逊棋蒙特卡洛代码c++

以下是一个基于蒙特卡洛树搜索算法的亚马逊棋 AI 代码，使用 C++ 实现：

#include <bits/stdc++.h>
using namespace std;
const int N = 10, M = 100000;
const int dir[8][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}};
struct Move {
    int x, y, nx, ny, cnt;
};
int n, m, t, a[N][N], f[N][N][N][N], vis[N][N], st[N][N];
int q[N * N][2], hh, tt, dx[M], dy[M], cnt[M];
int dist(int x, int y, int nx, int ny) {
    if (x == nx && y == ny) return 0;
    if (x == nx || y == ny || abs(x - nx) == abs(y - ny)) return max(abs(x - nx), abs(y - ny));
    return 2 * max(abs(x - nx), abs(y - ny));
}
int get(int x, int y) { return x * m + y; }
vector<Move> get_moves(int player) {
    vector<Move> res;
    memset(vis, 0, sizeof vis);
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j)
            if (a[i][j] == player) {
                for (int k = 0; k < 8; ++k) {
                    int nx = i + dir[k][0], ny = j + dir[k][1];
                    if (nx < 0 || nx >= n || ny < 0 || ny >= m || a[nx][ny]) continue;
                    memset(st, 0, sizeof st);
                    int hh = 0, tt = -1;
                    q[++tt][0] = i, q[tt][1] = j, st[i][j] = 1;
                    while (hh <= tt) {
                        int x = q[hh][0], y = q[hh++][1];
                        for (int l = 0; l < 8; ++l) {
                            int nx = x + dir[l][0], ny = y + dir[l][1];
                            if (nx < 0 || nx >= n || ny < 0 || ny >= m || st[nx][ny]) continue;
                            int d = dist(i, j, nx, ny);
                            if (d >= t) continue;
                            if (a[nx][ny] == 3 - player) {
                                tt = -1;
                                break;
                            }
                            q[++tt][0] = nx, q[tt][1] = ny, st[nx][ny] = 1;
                        }
                    }
                    if (tt == -1) continue;
                    for (int l = 0; l <= tt; ++l)
                        if (!vis[q[l][0]][q[l][1]]) {
                            vis[q[l][0]][q[l][1]] = 1;
                            res.push_back({i, j, q[l][0], q[l][1], tt + 1});
                        }
                }
            }
    return res;
}
int dfs(int player, int depth) {
    if (depth == 0) return 0;
    auto moves = get_moves(player);
    int res = -1;
    for (auto move : moves) {
        int x = move.x, y = move.y, nx = move.nx, ny = move.ny, cnt = move.cnt;
        a[nx][ny] = player, a[x][y] = 0;
        int t = dfs(3 - player, depth - 1);
        a[x][y] = player, a[nx][ny] = 0;
        if (t == -1) return -1;
        res = max(res, cnt - t);
    }
    return res;
}
void init() {
    memset(f, -1, sizeof f);
    queue<int> q;
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j)
            if (a[i][j] != 0) {
                f[i][j][i][j] = 0;
                q.push(get(i, j));
            }
    while (q.size()) {
        int t = q.front();
        q.pop();
        int x = t / m, y = t % m;
        for (int k = 0; k < 8; ++k) {
            int nx = x + dir[k][0], ny = y + dir[k][1];
            if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;
            if (f[x][y][nx][ny] == -1) {
                f[x][y][nx][ny] = f[x][y][x][y] + 1;
                q.push(get(nx, ny));
            }
        }
    }
}
int eval() {
    vector<int> p1, p2;
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j)
            if (a[i][j] == 1)
                p1.push_back(i * m + j);
            else if (a[i][j] == 2)
                p2.push_back(i * m + j);
    int res = 0;
    for (int i = 0; i < p1.size(); ++i)
        for (int j = 0; j < p2.size(); ++j)
            res += f[p1[i] / m][p1[i] % m][p2[j] / m][p2[j] % m];
    return res;
}
int dfs2(int player, int depth, int lf, int rf) {
    if (depth == 0) return eval();
    auto moves = get_moves(player);
    if (moves.empty()) return lf - rf;
    int res = -1e9;
    vector<int> cnts;
    for (auto move : moves) {
        int x = move.x, y = move.y, nx = move.nx, ny = move.ny, cnt = move.cnt;
        a[nx][ny] = player, a[x][y] = 0;
        cnts.push_back(cnt);
        res = max(res, -dfs2(3 - player, depth - 1, -rf, -lf));
        a[x][y] = player, a[nx][ny] = 0;
        lf = max(lf, res - cnt);
        if (lf >= rf) break;
    }
    return res;
}
void MCTS(int player, int depth) {
    auto moves = get_moves(player);
    if (moves.empty()) return;
    int ucnt = moves.size();
    int sum = 0;
    for (int i = 0; i < ucnt; ++i) {
        int x = moves[i].x, y = moves[i].y, nx = moves[i].nx, ny = moves[i].ny, cnt = moves[i].cnt;
        a[nx][ny] = player, a[x][y] = 0;
        int t = dfs(3 - player, depth - 1);
        a[nx][ny] = 0, a[x][y] = player;
        if (t != -1) cnts[i] += cnt - t, sum += cnt - t;
    }
    double mx = -1e9;
    int idx = -1;
    for (int i = 0; i < ucnt; ++i) {
        double val = 1.0 * cnts[i] / sum + sqrt(log(cnt[i]) / cnts[i]);
        if (val > mx) mx = val, idx = i;
    }
    int x = moves[idx].x, y = moves[idx].y, nx = moves[idx].nx, ny = moves[idx].ny;
    a[nx][ny] = player, a[x][y] = 0;
}
int main() {
    cin >> n >> m >> t;
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j) cin >> a[i][j];
    init();
    int depth = 0;
    while (1) {
        auto moves = get_moves(1);
        if (moves.empty()) {
            puts("Black loses");
            break;
        }
        if (depth >= 3) {
            memset(cnt, 0, sizeof cnt);
            for (int i = 0; i < M; ++i) cnt[i] = 1;
            for (int i = 0; i < 100; ++i) {
                int t = dfs2(1, 3, -1e9, 1e9);
                if (t > 0) break;
                int idx = rand() % moves.size();
                int x = moves[idx].x, y = moves[idx].y, nx = moves[idx].nx, ny = moves[idx].ny;
                a[nx][ny] = 1, a[x][y] = 0;
            }
        } else {
            for (int i = 0; i < 10000; ++i) MCTS(1, 5);
            int mx = -1e9, idx = -1;
            for (int i = 0; i < moves.size(); ++i)
                if (cnts[i] > mx) mx = cnts[i], idx = i;
            int x = moves[idx].x, y = moves[idx].y, nx = moves[idx].nx, ny = moves[idx].ny;
            a[nx][ny] = 1, a[x][y] = 0;
        }
        ++depth;
        auto moves2 = get_moves(2);
        if (moves2.empty()) {
            puts("White loses");
            break;
        }
        int idx = rand() % moves2.size();
        int x = moves2[idx].x, y = moves2[idx].y, nx = moves2[idx].nx, ny = moves2[idx].ny;
        a[nx][ny] = 2, a[x][y] = 0;
    }
    return 0;
}

该代码实现了亚马逊棋的基本算法，包括：

- 蒙特卡洛树搜索算法；
- 最短路算法；
- 评估函数。

其中，`get_moves` 函数用于获取当前玩家所有合法的移动方式；`dfs` 函数用于在搜索树上搜索指定深度的节点，并返回该节点的得分（如果该节点无法到达终局状态，则返回 -1）；`init` 函数用于预处理出最短路；`eval` 函数用于评估当前局面的得分；`dfs2` 函数用于在搜索树上搜索指定深度的节点，并返回该节点的得分；`MCTS` 函数用于执行一次蒙特卡洛树搜索，并更新访问次数和得分。

c＋＋亚马逊棋蒙特卡洛代码

以下是一个简单的 C++ Monte Carlo Tree Search 实现，用于玩亚马逊棋游戏：

#include <bits/stdc++.h>
using namespace std;

const int N = 10, INF = 1e9;
const int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
const int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};

int n, m, turn, tot, cnt;
int a[N][N], f[N * N][N * N], pos[N * N][2], val[N * N], vis[N * N], res[N * N];
double c1 = 1.0, c2 = sqrt(2.0);
mt19937 rnd(time(NULL));

inline bool check(int x, int y) {
    return x >= 1 && x <= n && y >= 1 && y <= m && !a[x][y];
}

inline int calc(int x) {
    int ans = 0;
    for(int i = 0; i < 8; i++) {
        int nx = pos[x][0] + dx[i], ny = pos[x][1] + dy[i];
        while(check(nx, ny)) ans++, nx += dx[i], ny += dy[i];
    }
    return ans;
}

inline double calcval(int x) {
    return 1.0 * val[x] / (vis[x] + 1) + c1 * sqrt(log(tot) / (vis[x] + 1));
}

inline int expand(int x) {
    int ans = 0;
    for(int i = 0; i < 8; i++) {
        int nx = pos[x][0] + dx[i], ny = pos[x][1] + dy[i];
        while(check(nx, ny)) {
            ans++;
            f[x][++cnt] = nx * m + ny;
            val[cnt] = calc(cnt);
            nx += dx[i], ny += dy[i];
        }
    }
    return ans;
}

inline int getmax() {
    int ans = 1;
    double mx = -INF;
    for(int i = 1; i <= cnt; i++) {
        if(!vis[i]) return i;
        double val = calcval(i);
        if(val > mx) mx = val, ans = i;
    }
    return ans;
}

inline int simulate(int x) {
    int ans = 0;
    for(int i = 0; i < 100; i++) {
        int nx = pos[x][0], ny = pos[x][1];
        while(1) {
            int p = rnd() % 8, xx = nx + dx[p], yy = ny + dy[p];
            if(check(xx, yy)) {
                nx = xx, ny = yy;
                break;
            }
        }
        int t = nx * m + ny;
        if(vis[t] || t == x) continue;
        vis[t] = 1;
        int num = expand(t);
        ans += num;
    }
    return ans;
}

inline int getbest(int x) {
    int ans = 1;
    double mx = -INF;
    for(int i = 1; i <= cnt; i++) {
        if(!vis[i]) continue;
        double val = 1.0 * res[i] / vis[i] + c2 * sqrt(log(tot) / vis[i]);
        if(val > mx) mx = val, ans = i;
    }
    return ans;
}

inline void backpropagate(int x, int v) {
    while(x) {
        vis[x]++, res[x] += v, v = -v, x = f[x][0];
    }
}

inline void mcts() {
    int now = 1;
    while(1) {
        if(!expand(now)) break;
        int idx = getmax();
        int num = simulate(idx);
        backpropagate(idx, num);
        turn = -turn;
        tot++;
        now = getbest(now);
    }
}

int main() {
    cin >> n >> m;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            cin >> a[i][j];
            if(a[i][j] == 1) pos[++cnt][0] = i, pos[cnt][1] = j;
        }
    }
    mcts();
    int ans = 1, mx = -INF;
    for(int i = 1; i <= cnt; i++) {
        if(!vis[i]) continue;
        if(res[i] * turn > mx) mx = res[i] * turn, ans = i;
    }
    cout << pos[ans][0] << " " << pos[ans][1] << endl;
    return 0;
}

该代码实现了一个简单的 Monte Carlo Tree Search 算法来玩亚马逊棋游戏。具体来说，它首先根据当前局面扩展一个节点，然后选择一个未被访问的子节点进行模拟，得到一个评估值，最后将评估值回溯到从该节点到根节点的所有节点。根据 MCTS 算法中的公式，每个节点的价值由它的胜率和探索程度决定，这里的实现中使用了 UCB1 公式。