Codeforces Round #551 (Div. 2) A-E

A. Serval and Bus

• 算出每辆车会在什么时候上车， 取min即可

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define M 101
#define mmp make_pair
using namespace std;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
int ans[M], n, t; 
int main() {
    n = read(), t = read();
    for(int i = 1; i <= n; i++)
    {
        int a = read(), b = read();
        while(a < t) a += b;
        ans[i] = a;
    }
    int minn = 0x3e3e3e3e, pl = 0;
    for(int i = 1; i <= n; i++) if(ans[i] < minn) minn = ans[i], pl = i;
    cout << pl << "\n";
    return 0;
}

B. Serval and Toy Bricks

• 贪心每个位置假如能放， 就放横看数看的较小值

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define M 110
#define mmp make_pair
using namespace std;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
int a[M][M], h[M], l[M];
int main() {
    int x = read(), y = read(), z = read();
    for(int i = 1; i <= y; i++) h[i] = read();
    for(int i = 1; i <= x; i++) l[i] = read();
    for(int i = 1; i <= x; i++) for(int j = 1; j <= y; j++) a[i][j] = read();
    for(int i = 1; i <= x; i++) for(int j = 1; j <= y; j++) if(a[i][j]) a[i][j] = min(l[i], h[j]);
    for(int i = 1; i <= x; i++) {
        for(int j = 1; j <= y; j++) cout << a[i][j] << " ";
        cout << "\n";
    }
    return 0;
}

C. Serval and Parenthesis Sequence

• 转化成第一个括号一定要和最后一个括号匹配， 对于2到n-1位置的串进行构造即可

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define M 300010
#define mmp make_pair
using namespace std;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}
char s[M];
int n;
int main() {
    n = read();
    scanf("%s", s + 1);
    if(n & 1) return 0 * puts(":(");
    if(s[1] == ')' || s[n] == '(') return 0 * puts(":(");
    s[1] = '(', s[n] = ')';
    int tot = n - 2, a = 0, b = 0, c;
    for(int i = 2; i < n; i++) {
        if(s[i] == '(') a++;
        else if(s[i] == ')') b++;
        else c++;
    }
    if(a > tot / 2 || b > tot / 2) return 0 * puts(":(");
    a = tot / 2 - a;
    for(int i = 2; i < n; i++) {
        if(s[i] == '?') {
            if(a) s[i] = '(', a--;
            else s[i] = ')';
        }
    }
    int cnt = 0;
    for(int i = 2; i < n; i++) {
        if(s[i] == '(') cnt++;
        else cnt--;
        if(cnt < 0) return 0 * puts(":(");
    }
    puts(s + 1);
    return 0;
}

D. Serval and Rooted Tree

• 树形dp， 考虑dp每颗子树的影响， 最大值的话选择影响最小的， 最小值的话影响求和

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define M 300010
#define mmp make_pair
using namespace std;
int read() {
    int nm = 0, f = 1;
    char c = getchar();
    for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    return nm * f;
}

int n, ver[M], fa[M], k, f[M];
vector<int> to[M];
bool cmp(int a, int b) {
    return f[a] < f[b];
}

void dfs(int now) {
    for(int i = 0; i < to[now].size(); i++) {
        int vj = to[now][i];
        dfs(vj);
    }
    if(to[now].size() == 0) {
        f[now] = 1;
        return;
    }
    sort(to[now].begin(), to[now].end(), cmp);
    if(ver[now]) {
        f[now] = f[to[now][0]];
    } else {
        for(int i = 0; i < to[now].size(); i++) f[now] += f[to[now][i]];
    }
}

int main() {
    n = read();
    for(int i = 1; i <= n; i++) ver[i] = read();
    for(int i = 2; i <= n; i++) fa[i] = read(), to[fa[i]].push_back(i);
    for(int i = 2; i <= n; i++) if(to[i].size() == 0) k++;
    dfs(1);
    cout << k - f[1] + 1<< "\n";
    return 0;
}

E. Serval and Snake

• 先枚举找到边界， 然后二分找位置

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#define ll long long
#define M 101
#define mmp make_pair
using namespace std;
bool rev = false;
vector<int>row, col;
int ask(int x1, int y1, int x2, int y2) {
    if(rev) {
        swap(x1, y1);
        swap(x2, y2);
    }
    cout << "? " << x1 + 1 << ' ' << y1 + 1 << ' ' << x2 + 1 << ' ' << y2 + 1 << endl;
    int res;
    cin>>res;
    return res;
}
void answer(int x1, int y1, int x2, int y2) {
    if(rev) {
        swap(x1, y1);
        swap(x2, y2);
    }
    cout << "! " << x1 + 1 << ' ' << y1 + 1 << ' ' << x2 + 1 << ' ' << y2 + 1 << endl;
}
int main() {
    int n;
    cin>>n;
    for(int i = 0; i < n; i++) {
        int res = ask(i, 0, i, n - 1);
        if(res % 2 == 1)row.push_back(i);
        res = ask(0, i, n - 1, i);
        if(res % 2 == 1)col.push_back(i);
    }
    vector<pair<int, int> >ans;
    if(row.size() == 2 && col.size() == 2) {
        for(int x:row)for(int y:col) {
                if(ask(x, y, x, y) % 2 == 1) {
                    ans.push_back(mmp(x, y));
                }
            }
        answer(ans[0].first, ans[0].second, ans[1].first, ans[1].second);
        return 0;
    }
    if(row.size() == 0) {
        rev = true;
        swap(row, col);
    }
    int ok = n - 1, ng =  - 1;
    while(ok - ng>1) {
        int t = (ok + ng)/2;
        if(ask(0, 0, row[0], t) % 2 == 1)ok = t;
        else ng = t;
    }
    answer(row[0], ok, row[1], ok);
    return 0;
}