Week 9

1-1 小朱爱

签到题，按题意数一下即可，注意 \ 的输出要用 \\

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

void solve(){
	string str;
    getline(cin,str);
	int cnt=0;
	for(int i=0;i<str.size();i++){
		if(str[i]=='l'||str[i]=='I'){
			cnt++;
		}
	}
	for(int i=1;i<=cnt;i++){
		cout<<'\\';
	}
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    //cin >> t;
    while(t--){
        solve();
    }
    return 0;
}

1-2 括号完美匹配

签到，可以用 vector 模拟栈，遇到左括号放入，遇到右括号尝试匹配即可。注意最后栈为空才是真正匹配完

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

void solve(){
    int n; string s;
    cin >> s;
    vector<char> stk;
    for(auto c : s){
        if(c == '(' || c == '[') stk.push_back(c);
        else if(c == ')'){
            if(stk.empty() || stk.back() != '('){
                cout << "no" << '\n';
                return;
            }
            else stk.pop_back();
        }
        else if(c == ']'){
            if(stk.empty() || stk.back() != '['){
                cout << "no" << '\n';
                return;
            }
            else stk.pop_back();
        }
    }
    if(!stk.empty()){
        cout << "no" << '\n';
        return;
    }
    cout << "yes" << '\n';
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t = 1; cin >> t;
    while(t--) solve();
    
    return 0;
}

1-3 跳一跳 (升级版)

求最小跳跃次数，考虑 BFS，用 mp 记录某值的所有下标。用 vis1[i] 标记 i 的最小到达时间，同时还需要用 vis2 标记某值是否用过第三种转送方式

#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;

constexpr int inf = 0x3f3f3f3f;

void solve(){
    int n;
    cin >> n;
    vector<int> a(n);
    map<int, vector<int>> idx; // v , vector<int>
    for(int i = 0; i < n; i++) {
        cin >> a[i];
        idx[a[i]].push_back(i);
    }

    vector<int> vis1(n, inf);
    set<int> vis2; // x
    queue<pii> q;
    q.push({0, 0});
    while(!q.empty()){
        auto[x, y] = q.front(); q.pop(); // c++ 17 语法 , 不能用可替换成 first , second
        if(vis1[x] != inf) continue;
        vis1[x] = y;
        
        if(x == n-1) break;
        
        q.push({x+1, y+1}); // 1
        if(x) q.push({x-1, y+1}); // 2

        if(!vis2.count(a[x])){ // 3
            for(int i : idx[a[x]]){
                q.push({i, y+1});
            }
            vis2.insert(a[x]);
        }
    }

    cout << vis1[n-1] << '\n';

}

int main(){
    solve();
}

1-4 马走日

枚举马所有可以走的方式，BFS 暴搜即可

#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;

constexpr int inf = 0x3f3f3f3f;

int dx[8] = {1,2,2,1,-1,-2,-2,-1};
int dy[8] = {2,1,-1,-2,2,1,-1,-2};

struct Node{
    int x, y, step;
};

void solve(){
    int n, m, sx, sy, tx, ty;
    cin >> n >> m >> sx >> sy >> tx >> ty;

    vector<vector<int>> vis(n, vector<int>(m, 0));
    queue<Node> q;
    q.push({sx, sy, 0});
    vis[sx][sy] = 1;

    int ans = inf;

    while(!q.empty()){
        auto [x, y, step] = q.front(); q.pop();

        if(x == tx && y == ty){
            ans = step;
            break;
        }

        for(int i = 0; i < 8; i++){                                         
            int nx = x + dx[i];
            int ny = y + dy[i];
            if(nx >= 0 && nx < n && ny >= 0 && ny < m && !vis[nx][ny]){
                vis[nx][ny] = 1;
                q.push({nx, ny, step+1});
            }
        }
    }

    cout << ans << '\n';
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t = 1;
    // cin >> t;
    while(t--) solve();

    return 0;
}

1-5 小红的大蘑菇

直接暴力把非安全区域标记出来，然后正常跑 BFS 即可

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

constexpr int INF = 0x3f3f3f3f;
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};

struct Node{
    int x, y, d;
};

void solve(){
    int n, m;
    cin >> n >> m;
    vector<vector<char>> mat(n+2, vector<char>(m+2, '#'));
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= m;j++){
            cin >> mat[i][j];
        }
    }

    vector<vector<int>> bad(n+2, vector<int>(m+2));

    /* 暴力标记不安全区域 */
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= m;j++){
            if(mat[i][j] == '*'){
                for(int dx = -1; dx <= 1; dx++){
                    for(int dy = -1; dy <= 1; dy++){
                        if(abs(dx) + abs(dy) > 1) continue;
                        bad[i+dx][j+dy] = 1;
                    }
                }
            }
            if(mat[i][j] == '#'){
                for(int dx = -2; dx <= 2; dx++){
                    for(int dy = -2; dy <= 2; dy++){
                        if(i+dx < 0 || i+dx > n || j+dy < 0 || j + dy > m) continue;
                        if(abs(dx) + abs(dy) > 2) continue;
                        bad[i+dx][j+dy] = 1;
                    }
                }
            }
        }
    }

    /* 起点 / 终点检查 */
    if(bad[1][1] || bad[n][m]){
        cout << -1 << '\n';
        return;
    }

    /* BFS 走安全空地 */
    vector<vector<bool>> vis(n+2, vector<bool>(m+2));
    queue<Node> q;
    q.push({1, 1, 0});
    vis[1][1] = true;
    while(!q.empty()){
        auto [x,y,d] = q.front(); q.pop();
        if(x == n && y == m){
            cout << d << '\n';
            return;
        }
        for(int k = 0;k < 4;k++){
            int nx = x + dx[k];
            int ny = y + dy[k];
            if(vis[nx][ny] || bad[nx][ny] || mat[nx][ny] != '.') continue;
            vis[nx][ny] = true;
            q.push({nx,ny,d+1});
        }
    }
    cout << -1 << '\n';
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    solve();
}

1-6 回家

例题原题，在讲课 pdf 里面

#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;

int n, m;
struct man{
    int x, y, hp;
};

int bfs(auto& mat, int Sx, int Sy){
    queue<man> q;
    vector<vector<int>> vis(n+2, vector<int> (m+2, 0)); // 当 hp 大于原来, 则可以再走此格
    vector<pii> moves = {{-1,0},{1,0},{0,1},{0,-1}};
    q.push({Sx, Sy, 6});
    vis[Sx][Sy] = 6;
    
    int depth = 0; // 记录最短路径
    while(!q.empty()){
        int sz = q.size();
        for(int i=0; i<sz; i++){
            int cx = q.front().x;
            int cy = q.front().y;
            int chp = q.front().hp;
            q.pop();

            if(chp == 0) continue;

            if(mat[cx][cy] == 3) return depth; // 找到家直接 return
            if(mat[cx][cy] == 4){ // 捡到 鼠标? 回满hp
                chp = 6;
            }

            for(auto& mo : moves){
                int nx = mo.first + cx;
                int ny = mo.second + cy;
                int nhp = chp - 1;
                if(nhp <= vis[nx][ny] || mat[nx][ny] == 0) continue;
                vis[nx][ny] = nhp;
                q.push({nx, ny, nhp});
            }

        }
        depth++;
    }
    return -1;
}

int main(){
    int Sx, Sy;
    cin >> n >> m;
    vector<vector<int>> mat(n+2, vector<int>(m+2, 0));
    for(int r=1; r<=n; r++){
        for(int c=1; c<=m; c++){
            cin >> mat[r][c];
            if(mat[r][c] == 2){
                Sx = r;
                Sy = c;
            }
        }
    }
    int ans = bfs(mat, Sx, Sy);
    cout << ans << '\n';
}

1-7 Nightmare II

先跑一遍 npy 的 BFS，在跑一遍 erriyue 的 BFS，记录到某格的最短时间。或者说直接按时间模拟。可以计算曼哈顿距离，来计算鬼是否在某个时间扩散到某格

#include <bits/stdc++.h>
using namespace std;
using pii = pair<int,int>;

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};

void solve() {
    int n, m;
    cin >> n >> m;

    vector<vector<char>> mat(n + 2, vector<char>(m + 2, 'X'));
    vector<vector<bool>> vis1(n + 2, vector<bool>(m + 2)), vis2 = vis1;
    for (int i = 1; i <= n; i++) {
        string row; cin >> row;
        for (int j = 1; j <= m; j++) {
            mat[i][j] = row[j - 1];
        }
    }

    queue<pii> q1, q2;
    vector<pii> ghosts;
    for(int i=1; i<=n; i++){
        for(int j=1; j<=m ;j++){
            if(mat[i][j] == 'M') q1.push({i, j}); // 男
            if(mat[i][j] == 'G') q2.push({i, j}); // 女
            if(mat[i][j] == 'Z') ghosts.push_back({i, j}); // 鬼
        }
    }
    
    // 检查某个点在 time 秒时是否安全
    // 鬼在 time 秒的覆盖半径是 2 * time
    auto is_safe = [&](int x, int y, int time)->bool {
        for (auto z : ghosts) {
            if(abs(x - z.first) + abs(y - z.second) <= 2 * time){
                return false;
            }
        }
        return true;
    };

    // q: 当前人物的队列
    // steps: 这一秒可以走的步数 (Boy=3, Girl=1)
    // vis: 当前人物的访问数组
    int time = 0;
    auto bfs = [&](queue<pii> &q, int steps, vector<vector<bool>>& vis)->bool {
        for (int k = 0; k < steps; k++) {
            int sz = q.size(); // 当前层的节点数，保证只扩展一层，也就是一步一步走
            while (sz--) {
                int x = q.front().first, y = q.front().second;
                q.pop();

                if(!is_safe(x, y, time)){
                    continue;
                }

                for (int i = 0; i < 4; i++) {
                    int nx = x + dx[i];
                    int ny = y + dy[i];

                    if (!is_safe(nx, ny, time) || mat[nx][ny]=='X') continue;
                    if (vis[nx][ny]) continue;
                    vis[nx][ny] = true;

                    if(vis1[nx][ny] && vis2[nx][ny]) return true; // 两人相遇

                    q.push({nx, ny});
                }
            }
        }
        return false;
    };

    // 时间模拟
    // 只要还有人能动，就继续模拟
    while (!q1.empty() || !q2.empty()) {
        time++;

        // Erriyue 动 3 步
        if (bfs(q1, 3, vis1)) {
            cout << time << '\n';
            return;
        }

        // Girl 动 1 步
        if (bfs(q2, 1, vis2)) {
            cout << time << '\n';
            return;
        }
    }
    cout << -1 << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    if (!(cin >> t)) return 0;
    while (t--) solve();
    return 0;
}

1-8 连通块的权

裸 BFS，也可以直接 DFS

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

using ll = long long;
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};

void solve(){
    int n, m;
    cin >> n >> m;

    vector<vector<int>> mat(n+2, vector<int>(m+2));
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            cin >> mat[i][j];
        }
    }

    vector<vector<bool>> vis(n+2, vector<bool>(m+2, false));
    ll ans = 0;

    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(mat[i][j] == 0 || vis[i][j]) continue;

            ll sum = 0;
            queue<pair<int,int>> q;
            q.push({i, j});
            vis[i][j] = true;

            while(!q.empty()){
                auto [x, y] = q.front(); q.pop();
                sum += mat[x][y];

                for(int k = 0; k < 4; k++){
                    int nx = x + dx[k];
                    int ny = y + dy[k];
                    if(vis[nx][ny] || mat[nx][ny] == 0) continue;
                    vis[nx][ny] = true;
                    q.push({nx, ny});
                }
            }
            ans = max(ans, sum);
        }
    }

    cout << ans << '\n';
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;
    cin >> T;
    while(T--) solve();
}

1-9 选数

暴搜即可，这道题 DFS，BFS 实现都行，本质没什么区别。每个位置选或不选，时间复杂度 \(O(2^n)\)

BFS 实现

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

using ll = long long;

bool isprime(int x){
    if(x <= 1) return false;
    for(int i = 2; i * i <= x; i++)
        if(x % i == 0) return false;
    return true;
}

struct Node{
    int cnt;   // 已选元素数量
    int sum;   // 当前和
    int pos;   // 下一轮可以选择的起始下标
};

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for(int i = 0; i < n; i++) cin >> a[i];

    ll ans = 0;
    queue<Node> q;
    q.push({0, 0, 0});

    while(!q.empty()){
        auto [cnt, sum, pos] = q.front(); q.pop();
        if(cnt == k){
            if(isprime(sum)) ans++;
            continue;
        }
        for(int i = pos; i < n; i++){
            q.push({cnt + 1, sum + a[i], i + 1});
        }
    }

    cout << ans << '\n';
}

DFS 实现

#include <iostream>
#include <cstdio>
using namespace std;
bool isprime(int a){
    if(a == 1) return false;
    for(int i = 2; i * i <= a; i++)
        if(a % i == 0)
            return false;
    return true;
}

int n,k;
int a[25];
long long ans;

void dfs(int m, int sum, int startx){
    if(m == k){
        if(isprime(sum))
            ans++;
        return ;
    }
    for(int i = startx; i < n; i++)
        dfs(m + 1, sum + a[i], i + 1);
    return ;
}

int main(){
    scanf("%d%d",&n,&k);
    for(int i = 0; i < n; i++)
        scanf("%d",&a[i]);
    dfs(0, 0, 0);
    printf("%d\n",ans);
    return 0;
}

1-10 夺宝大赛

题解

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

#define ios ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
typedef pair<int, int> pii;
const int N = 110, M = 5010;
const int dx[4] = {1, -1, 0, 0}, dy[4] = {0, 0, 1, -1};

int n, m, q, sx, sy, g[N][N], d[N][N];
pii des[M];

void bfs(){
    queue<pii> q;
    q.emplace(sx, sy);
    d[sx][sy] = 0;

    while(q.size()){
        auto [x, y] = q.front();
        q.pop();

        for(int k = 0; k < 4; k ++ ){
            int nx = x + dx[k], ny = y + dy[k];
            if(nx < 1 || ny < 1 || nx > n || ny > m) continue;
            if(!g[nx][ny] || ~d[nx][ny]) continue;
            q.emplace(nx, ny);
            d[nx][ny] = d[x][y] + 1;
        }
    }
}

int main(){
    ios;

    cin >> n >> m;
    memset(d, -1, sizeof d);
    for(int i = 1; i <= n; i ++ )
        for(int j = 1; j <= m; j ++ ){
            cin >> g[i][j];
            if(g[i][j] == 2) sx = i, sy = j;
        }

    cin >> q;
    for(int i = 1, x, y; i <= q; i ++ ) cin >> x >> y, des[i] = {y, x};

    bfs();

    unordered_map<int, int> cnt; // 每个距离有多少个队伍
    for(int i = 1; i <= q; i ++ ){
        auto [x, y] = des[i];
        cnt[d[x][y]] ++ ;
    }
 
    int mn = 1e9, ans = -1;
    for(int i = 1; i <= q; i ++ ){
        auto [x, y] = des[i];
        if(d[x][y] == -1 || cnt[d[x][y]] > 1) continue;
        if(d[x][y] < mn) mn = d[x][y], ans = i;
    }

    if(ans == -1) cout << "No winner." << "\n";
    else cout << ans << ' ' << mn << "\n";

    return 0;
}