Week5

蓝桥杯模拟赛 3

A题 特殊年份

签到题

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long

using namespace std;

typedef double D;
const int N = 10010, M = 1e9 + 7;

int ans;

void solve() {
    for (int i = 0; i < 5; i++) {
        int n;
        cin >> n;
        if ((n / 1000 == n % 100 / 10) && (n / 100 % 10 + 1 == n % 10))
            ans++;
    }
    cout << ans;
}
int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--)solve();
    return 0;
}

B题 小平方

按照题目意思模拟一遍就可以了，但要小心除而会向零取整，强制转化成浮点数double

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long

using namespace std;

typedef double D;
const int N = 10010, M = 1e9 + 7;

int ans;

void solve() {
    int n;
    cin >> n;
    for (int i = 1; i < n; i++) {
        int res = (i * i) % n;
        if (res < (D) n / 2 )ans++;
    }
    cout << ans;
}
int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--)solve();
    return 0;
}

C题 砝码称重

dp问题，使用集合的思想来做，f[i,j]的含义为前i个物品总重量不超过j的选的方法!，以最后一次状态来分析，分为不拿砝码f[i-1,j]，加到左边f[i-1,j-w]，加到右边f[i-1,j+w]，这三种状态只要一个非空，就非空，但是出现了j-w有可能是负数，所以可以给所有的j加一个偏移量，或者使用镜像解题f[i-1,abs(j-w)]，初始状态f[0][0]=1。

偏移量，防止数组下标越界。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

int w[110];
bool f[110][2*N];
int n, sum;

void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++)cin >> w[i], sum += w[i];
    f[0][N] = true;
    for (int i = 1; i <= n; i++) {
        for (int j = -sum; j <= sum; j++) {
            f[i][j + N] = f[i - 1][j + N];
            if (j - w[i] >= -sum)f[i][j + N] |= f[i - 1][j - w[i] + N];
            if (j + w[i] <= sum)f[i][j + N] |= f[i - 1][j + w[i] + N];
        }
    }
    int res = 0;
    for (int i = 1; i < N * 2; i++)
        if (f[n][i + N] == 1)res++;
    cout << res << endl;
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

镜像运用绝对值

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

int w[110];
bool f[110][2*N];
int n, sum;

void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++)cin >> w[i], sum += w[i];
    f[0][0] = true;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= sum; j++) {
            f[i][j] = f[i - 1][j] || f[i - 1][j + w[i]] || f[i - 1][abs(j - w[i])];
        }
    }
    int res = 0;
    for (int i = 1; i < N * 2; i++)
        if (f[n][i])res++;
    cout << res << endl;
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

如果暴力来做会超时，dfs，只有一半的分数

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long

using namespace std;

typedef double D;
const int N = 110, M = 1e5 + 10;

int a[N], st[M];
int n, sum;

void dfs(int s,int u) {
    if (u > n) {
        if(s>0)st[s]=1;
        return ;
    }
    dfs(s + a[u], u + 1);
    dfs(s, u + 1);
    dfs(abs(s - a[u]), u + 1);
}

void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++)cin >> a[i],sum+=a[i];
    dfs(0, 1);
    int ans = 0;
    for (int i = 1; i < M; i++) {
        if (st[i])ans++;
    }
    cout << ans << endl;
}
int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

D题 完全平方数

不能暴力做，会超时，使用分解质因的方式求答案，一个完全平方数的质因子的次数一定是偶数，求一个数最小乘以多少是完全平方数，就求这个数的质因子的次数，如果是偶数就不成，如果是奇数就乘以质因子，次数就变成偶数了，最后在判断剩余的n是不是大于一，如果是还有一个质因子的次数是奇数，再乘以它就是答案。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

ll n;

void solve() {
    cin >> n;
    ll res = 1;
    for (ll i = 2; i * i <= n; i++) {
        int cnt = 0;
        if (n % i == 0) {
            while (n % i == 0)cnt++, n /= i;
            if (cnt & 1)res *= i;
        }
    }
    if (n > 1)res *= n;
    cout << res << endl;
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

E题 时间显示

注意单位是毫秒，先转化成秒，将一整天所需要的秒去掉，再一个一个求小时，分钟，秒。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long

using namespace std;

typedef double D;
const int N = 10010, M = 1e9 + 7;

int ans;

void solve() {
    ll n;
    cin>>n;
    n/=1000;
    ll day=24*60*60;
    int k=n%day;
    ll s=k%60;
    ll h=k/3600;
    ll m=(k-h*3600)/60;
    printf("%02lld:%02lld:%02lld",h,m,s);
}
int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--)solve();
    return 0;
}

H题 负载均衡

对每个电脑工作结束的时间，用一个对来进行维护，把每次结束时间再开始时间之前的电脑都删掉计算算力。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 2e5 + 10, M = 1e5 + 10;

priority_queue<PII,vector<PII>,greater<PII>> q[N];
int n,m;
int aa[N];

void solve() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++)cin >> aa[i];
    while (m--) {
        int a, b, c, d;
        cin >> a >> b >> c >> d;
        while (q[b].size() && q[b].top().first <= a) {
            aa[b] += q[b].top().second;
            q[b].pop();
        }
        if (aa[b] < d)cout << "-1" << endl;
        else {
            q[b].push({a + c, d});
            aa[b] -= d;
            cout << aa[b] << endl;
        }
    }
}
int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

单链表

单链表的复习，建图的基础。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

int h=-1,e[N],ne[N],idx=0;

void add_head(int x) {
    e[idx] = x;
    ne[idx] = h;
    h = idx++;
}

void add(int k,int x) {
    e[idx] = x;
    ne[idx] = ne[k];
    ne[k] = idx++;
}

void remove(int k) {
    ne[k] = ne[ne[k]];
}

void solve() {
    int m;
    cin >> m;
    while (m--) {
        char ch;
        cin >> ch;
        if (ch == 'H') {
            int x;
            cin >> x;
            add_head(x);
        } else if (ch == 'D') {
            int k;
            cin >> k;
            if (!k)h = ne[h];
            remove(k - 1);
        } else {
            int k, x;
            cin >> k >> x;
            add(k - 1, x);
        }
    }
    for (int i = h; i != -1; i = ne[i])cout << e[i] << ' ';
    cout << endl;
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

模拟栈

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

int m, tt;
int stk[N];

void solve() {
    cin >> m;
    while (m--) {
        string s;
        int n;
        cin >> s ;
        if (s == "push") {
            cin >> n;
            stk[++tt] = n;
        }
        else if (s == "pop") tt--;
        else if (s == "empty") {
            if (tt == 0)cout << "YES" << endl;
            else cout << "NO" << endl;
        } else cout << stk[tt] << endl;
    }
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

模拟队列

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

int m, tt, hh;
int q[N];

void solve() {
    cin >> m;
    while (m--) {
        string s;
        int n;
        cin >> s;
        if (s == "push") {
            cin >> n;
            q[tt++] = n;
        } else if (s == "pop") hh++;
        else if (s == "empty") {
            if (hh >= tt)cout << "YES" << endl;
            else cout << "NO" << endl;
        } else cout << q[hh] << endl;
    }
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

双链表

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

int idx,e[N],l[N],r[N];

void init(){
    l[1]=0,r[0]=1,idx=2;
}

void add(int k,int x) {
    e[idx] = x;
    l[idx] = k, r[idx] = r[k];
    l[r[k]] = idx, r[k] = idx++;
}

void remove(int k) {
    r[l[k]] = r[k];
    l[r[k]] = l[k];
}

void solve() {
    init();
    int m;
    cin >> m;
    while (m--) {
        string op;
        int k, x;
        cin >> op;
        if (op == "L") {
            cin >> x;
            add(0, x);
        } else if (op == "R") {
            cin >> x;
            add(l[1], x);
        } else if (op == "D") {
            cin >> k;
            remove(k + 1);
        } else if (op == "IR") {
            cin >> k >> x;
            add(k + 1, x);
        } else {
            cin >> k >> x;
            add(l[k + 1], x);
        }
    }
    for (int i = r[0]; i != 1; i = r[i])cout << e[i] << ' ';
    cout << endl;
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

SMU Winter 2023 Round #12 (Div.2)

A题 kk画猪

签到题

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

void solve() {
    string s;
    cin >> s;
    printf(R"(  (\____/)
  / @__@ \
 (  (oo)  )
  `-.~~.-'
   /    \
 @/      \_
(/ /    \ \)
 WW`----'WW
)");
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

B题 kk学几何

签到题

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

int fa[N],a[N];

void solve() {
    int T;
    cin >> T;
    double ans = 0;
    while (T--) {
        int n;
        cin >> n;
        if (n == 1) {
            int r;
            cin >> r;
            ans += 3 * r * r;
        } else if (n == 2) {
            int l, h;
            cin >> l >> h;
            ans += (D) (l * h) / 2;
        } else if (n == 3) {
            int l, w;
            cin >> l >> w;
            ans += l * w;
        }
    }
    printf("%.1f\n", ans);
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

C题 kk算日期

本题要注意题目，公元0年不存在，看是看见了，没转过弯来，公元0年不存在，那么1后面就是-1，所以当所求公元年数小于等于0时，要再减1，再求绝对值计算即可。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

int fa[N],a[N];

void solve() {
    int n, m;
    cin >> n >> m;
    n += m;
    if (n <= 0)n--;
    n = abs(n);
    if (n % 4 == 0 && n % 100 != 0 || n % 400 == 0)
        cout << 29 << endl;
    else cout << 28 << endl;
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    cin >> h_h;
    while (h_h--) solve();
    return 0;
}

G题 kk看跳舞

问给出的数字首尾连接在一起能不能构成一个依次增加1或减少1的全排列。我们可以先找起始位置1位置然后再从两边判断，哪边是二就往哪边走，遍历到头或尾的时候更新成尾或头，直到遍历到全排列的最后一个数字，并计数cnt，看看cnt是不是和全排列最后一个数字相等，是就可以，不是就不可以。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 10010, M = 1e5 + 10;

int fa[N],a[N];

void solve() {
    int n;
    cin >> n;
    int start;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        if (a[i] == 1)start = i;
    }
    int cnt = 1;
    if ((a[start - 1] == 2 && start - 1 >= 1) || (start - 1 < 1 && a[n] == 2)) {
        while (1) {
            cnt++;
            if (start - 1 < 1)start = n + 1;
            if (a[--start] == n)break;
        }
    } else if ((a[start + 1] == 2 && start + 1 <= n) || (start + 1 > n && a[1] == 2)) {
        while (1) {
            cnt++;
            if (start + 1 == n + 1)start = 0;
            if (a[++start] == n)break;
        }
    }
    if (cnt == n)cout << "YES" << endl;
    else cout << "NO" << endl;
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    cin >> h_h;
    while (h_h--) solve();
    return 0;
}

H题 kk与十佳

要考虑所以的情况，所有是正数，所有是负数，负数和正数都有，有负数情况下还要考虑负数的奇偶。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 30010, M = 1e5 + 10;

vector<int>a,b;

void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        if (x > 0)a.push_back(x);
        else b.push_back(x);
    }
    sort(a.begin(), a.end());
    sort(b.begin(), b.end());
    if (a.size() == 0) {
        if (b.size() % 2 == 0)cout << b[0];
        else cout << b[b.size() - 1];
    } else if (b.size() == 0)cout << a[0];
    else {
        if (b.size() % 2 != 0)cout << b[b.size() - 1];
        else cout << a[0];
    }
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

I题 kk买股票

本题数据范围是1e5，用两重循环显然会超时，所以可以使用双指针来维护区间，我们保证区间的头是最小的，区间的尾是最大的，所以没个数都判断一下更新左右两端，并且左端点不能大于等于右端点，当大于时，要更新右端点。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

int n;
int a[N];

void solve() {
    cin >> n;
    for (int i = 0; i < n; i++)cin >> a[i];
    int mn = 1000000, mx = 0;
    int x, y;
    for (int i = 1, j = 0; j < n;) {
        if (a[i] > mx && i < n)mx = a[i], i++, x = i;
        else i++;
        if (a[j] < mn)mn = a[j], j++, y = j;
        else j++;
        if (j >= i)mx = a[j + 1], i = j + 1;
    }
    if (x - y <= 0)cout << 0 << endl;
    else cout << mx - mn << endl;
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

J题 kk与英语

就是简单的字符串模拟,注意读入有空格的字符串用scanf或getline。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e5 + 10;

int n;
int a[N];

void solve() {
    string s;
    getline(cin, s);
    cout << s[0];
    for (int i = 1; i < s.size(); i++) {
        if (s[i] == 'i' && s[i + 1] == 's' && s[i - 1] == ' ' && s[i + 2] == ' ') {
            cout << "was";
            i++;
            continue;
        }
        cout << s[i];
    }
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

KMP字符串

讲述怎么使用时间复杂度更低的方法寻找子串，感觉使用了双指针的算法。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<stack>
#include<map>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 100010, M = 1e6 + 10;

int n, m;
char s[M], p[N];
int ne[N];
map<string,int> mp;
set<string>st;

void solve() {
    cin >> n >> p + 1 >> m >> s + 1;

    for (int i = 2, j = 0; i <= n; i++) {
        while (j && p[i] != p[j + 1]) j = ne[j];
        if (p[i] == p[j + 1]) j++;
        ne[i] = j;
    }

    for (int i = 1, j = 0; i <= m; i++) {
        while (j && s[i] != p[j + 1]) j = ne[j];
        if (s[i] == p[j + 1]) j++;
        if (j == n) {
            printf("%d ", i - n);
            j = ne[j];
        }
    }
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

献给()阿尔吉侬的花束

最简单的bfs的题目，好像dfs也可以。，不可以求最短距离，权值为1。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 210, M = 1e5 + 10;

int n,m;
char g[N][N];
int dist[N][N];
bool vis[N][N];
int dx[4]={-1,1,0,0};
int dy[4]={0,0,1,-1};

int bfs(PII Start,PII End) {
    queue<PII> q;
    //memset(dist, -1, sizeof dist);
    q.push(Start);
    dist[Start.f][Start.s] = 0;
    vis[Start.f][Start.s] = true;
    while (!q.empty()) {
        auto t = q.front();
        q.pop();
        for (int i = 0; i < 4; i++) {
            int xx = dx[i] + t.f, yy = dy[i] + t.s;
            if (xx < 0 || xx >= n || yy < 0 || yy >= m)continue;
            if (vis[xx][yy])continue;
            if (g[xx][yy] == '#')continue;
            dist[xx][yy] = dist[t.f][t.s] + 1;
            if (End == make_pair(xx, yy))return dist[xx][yy];
            q.push({xx, yy});
            vis[xx][yy] = true;
        }
    }
    return -1;
}

void solve() {
    int T;
    cin >> T;
    while (T--) {
        memset(dist, 0, sizeof dist);
        memset(vis, false, sizeof vis);
        cin >> n >> m;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                cin >> g[i][j];
        PII start, end;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (g[i][j] == 'S')start = (make_pair(i, j));
                else if (g[i][j] == 'E')end = (make_pair(i, j));
            }
        }
        int dis = bfs(start, end);
        if (dis == -1)cout << "oop!" << endl;
        else cout << dis << endl;
    }
}

int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

红与黑

Flood fill模型

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#define IOS ios::sync_with_stdio(false)
#define CC cin.tie(nullptr),cout.tie(nullptr)
#define ll long long
#define f first
#define s second

using namespace std;

typedef pair<int ,int> PII;
typedef double D;
const int N = 210, M = 1e5 + 10;

int n,m;
char g[N][N];
int dist[N][N];
bool vis[N][N];
int dx[4]={-1,1,0,0};
int dy[4]={0,0,1,-1};

int dfs(int x,int y) {
    int cnt = 1;
    vis[x][y] = true;
    for (int i = 0; i < 4; i++) {
        int xx = x + dx[i], yy = y + dy[i];
        if (xx >= 0 && xx < n && yy >= 0 && yy < m && !vis[xx][yy] && g[xx][yy] == '.')
            cnt += dfs(xx, yy);
    }
    return cnt;
}

void solve() {
    while (cin >> m >> n) {
        if (n == 0 && m == 0)break;
        memset(vis, false, sizeof vis);
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                cin >> g[i][j];
        int x,y;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (g[i][j] == '@')
                    x=i,y=j;
        cout << dfs(x,y) << endl;
    }
}
int main() {
    IOS;
    CC;
    int h_h = 1;
    //cin >> h_h;
    while (h_h--) solve();
    return 0;
}

__EOF__

本文作者：north_h
本文链接：https://www.cnblogs.com/north-h/p/17619571.html
关于博主：评论和私信会在第一时间回复。或者直接私信我。
版权声明：本博客所有文章除特别声明外，均采用 BY-NC-SA 许可协议。转载请注明出处！
声援博主：如果您觉得文章对您有帮助，可以点击文章右下角【推荐】一下。您的鼓励是博主的最大动力！
posted @   north_h  阅读(18)  评论(0)  编辑  收藏  举报  
相关博文：
·  Week6
·  Week2
·  week6
·  2023寒假训练week4
·  第二周总结报告
阅读排行：
· TypeScript + Deepseek 打造卜卦网站：技术与玄学的结合
· 阿里巴巴 QwQ-32B真的超越了 DeepSeek R-1吗？
· 如何调用 DeepSeek 的自然语言处理 API 接口并集成到在线客服系统
· 【译】Visual Studio 中新的强大生产力特性
· 2025年我用 Compose 写了一个 Todo App
点击右上角即可分享
微信分享提示
SQL AI 助手