acwing 資料結構

陣列模擬單鏈表

int idx, head, ne[N5], e[N5];
int main() {
    int n = read(), k, x;
    head = -1; //初始化頭結點
    char op;
    while (n--) {
        scanf("%c", &op);
        k = read();
        if (op == 'H') {
            e[idx] = k;
            ne[idx] = head;
            head = idx++;
        }
        else if (op == 'D') {
            if (k == 0) { head = ne[head]; }
            else ne[k - 1] = ne[ne[k - 1]];
        }
        else {
            x = read();
            ne[idx] = ne[k - 1];
            ne[k - 1] = idx;
            e[idx++] = x;
        }
    }
    while (head != -1) {
        O(e[head]);
        head = ne[head];
    }
    return 0;
}
 

陣列模擬雙鏈表

int idx, head, l[N5], e[N5], r[N5];
void init() {
    r[0] = 1, l[1] = 0, idx = 2;
}
void insert(int k, int x) {
    l[idx] = k, r[idx] = r[k], e[idx] = x;
    l[r[k]] = idx, r[k] = idx++;
}
void remove(int k) {
    r[l[k]] = r[k], l[r[k]] = l[k];
}
int main() {
    init();
    int k, x, m = read();
    string op;
    while (m--) {
        cin >> op;k = read();
        if (op[0] == 'R') insert(l[1], k);
        else if (op[0] == 'L') insert(0, k);
        else if (op[0] == 'I') {
            x = read();
            if (op[1] == 'R') insert(k + 1, x);
            else insert(l[k + 1], x);
        }
        else  remove(k + 1);
    }
    head = r[head];
    while (head != 1) O(e[head]), head = r[head];
    return 0;
}
 

模擬棧

雪菜有的是 從 tt=0 不知道有沒有影響.

然後empty() 是 tt>0

int stk[N5], tt = -1; //記住下標是從 -1 開始的
void push(int x) {
    stk[++tt] = x;
}
void pop() {
    --tt;
}
int query() {
    return stk[tt];
}
bool empty() {
    return tt == -1;
}

模擬佇列

int q[N5], tt = -1, hh;
void push(int x) {
    q[++tt] = x;
}
void pop() {
    ++hh;
}
int query() {
    return q[hh];
}
bool empty() {
    return hh > tt;
}
 

單調棧

int a[N6], tt;
int main() {
    int n = read(), x;
    rep(i, 0, n) {
        x = read();
        while (tt && a[tt] >= x) tt--;
        if (tt) O(a[tt]);
        else O(-1);
        a[++tt] = x;
    }
    return 0;
}

滑動視窗最大值和最小值.

int a[N6], q[N6], tt = -1, hh;
int main() {
    int n = read(), k = read();
    rep(i, 0, n) a[i] = read();
    rep(i, 0, n) {
        if (hh <= tt && i - k + 1 > q[hh]) hh++;
        while (hh <= tt && a[q[tt]] >= a[i]) tt--; //最小值, 單增區間
        q[++tt] = i;
        if (i - k + 1 >= 0) O(a[q[hh]]);
    }
    puts("");
    tt = -1, hh = 0;
    rep(i, 0, n) {
        if (hh <= tt && i - k + 1 > q[hh]) hh++;
        while (hh <= tt && a[q[tt]] <= a[i]) tt--; //最大值, 單減區間
        q[++tt] = i;
        if (i - k + 1 >= 0) O(a[q[hh]]);
    }
    return 0;
}

表示式求值

不會做... 不過這個順序..

stack<int> op;
stack<int> num;
void eval() {
    auto b = num.top(); num.pop();
    auto a = num.top(); num.pop();
    auto c = op.top();op.pop();
    int x;
    if (c == '+') x = a + b;
    else if (c == '-') x = a - b;
    else if (c == '*') x = a * b;
    else x = a / b;
    num.push(x);
}
int main() {
    unordered_map<char, int> pri = { {'+',1},{'-',1},
            {'*',2},{'/',2} };
    string equ;
    cin >> equ;
    for (int i = 0;i < equ.size();i++) {
        auto c = equ[i];
        if (isdigit(c)) {
            int x = 0, j = i;
            while (j < equ.size() && isdigit(equ[j]))
                x = x * 10 + equ[j++] - '0';
            i = j - 1; num.push(x);
        }
        else if (c == '(') op.push(c);
        else if (c == ')') {
            while (op.top() != '(') eval();
            op.pop();
        }
        else {
            while (op.size() && op.top() != '(' &&
                pri[op.top()] >= pri[c]) eval();
            op.push(c);
        }
    }
    while (op.size()) eval();
    cout << num.top() << endl;
    return 0;
}