Codeforces Round #698 (Div. 2)

E. Nezzar and Binary String

题意：

两个(01)串(s)，(f)，(q)个询问区间，每次询问之后可以修改询问的区间内不超过一半的字符，问能否做到每次询问的区间内字符相同（全(1)或全(0)）并且所有询问结束后(s)与(f)相同

思路：

倒着做，线段树维护区间赋值

#include<bits/stdc++.h>
#define ls rt<<1
#define rs rt<<1|1
using namespace std;
typedef long long ll;
const int maxn = 200010;
char s[maxn], f[maxn];
int l[maxn], r[maxn];
int sum[maxn << 2], lazy[maxn << 2];
void pushUp(int rt){
    sum[rt] = sum[ls] + sum[rs];
}
void build(int rt, int l, int r){
    lazy[rt] = -1;
    if(l==r){
        sum[rt] = f[l]-'0';
        return ;
    }
    int mid = (l + r) >> 1;
    build(ls,l,mid);
    build(rs,mid+1,r);
    pushUp(rt);
}
void pushDown(int rt, int l, int r){
    if(lazy[rt]!=-1){
        lazy[ls] = lazy[rs] = lazy[rt];
        int mid = (l + r) >> 1;
        sum[ls] = lazy[ls] * (mid - l + 1);
        sum[rs] = lazy[rs] * (r - mid);
        lazy[rt] = -1;
    }
}
void update(int rt, int l, int r, int p, int L, int R){
    if(L<=l&&r<=R){
        sum[rt] = (r - l + 1) * p;
        lazy[rt] = p;
        return ;
    }
    pushDown(rt, l, r);
    int mid = (l + r) >> 1;
    if(mid >= L)
        update(ls, l, mid, p, L, R);
    if(mid < R)    
        update(rs, mid + 1, r, p, L, R);
    pushUp(rt);
}
int query(int rt, int l, int r, int L, int R){
    if(L<=l&&r<=R)return sum[rt];
    pushDown(rt, l, r);
    int mid = (l + r) >> 1, ans = 0;
    if(mid >= L)
        ans += query(ls, l, mid, L, R);
    if(mid < R)
        ans += query(rs, mid + 1, r, L, R);
    return ans;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t, n, q;
    cin >> t;
    while(t--) {
        cin>> n >> q >> (s+1) >> (f+1);
        for(int i = 0; i < q; i++) cin >> l[i] >> r[i];
        reverse(l, l + q);
        reverse(r, r + q);
        build(1, 1, n);
        int flag = 0;
        for(int i = 0; i < q; i++) {
            int res = query(1, 1, n, l[i], r[i]);
            if((r[i] - l[i] + 1) % 2 == 0 && res == (r[i] - l[i] + 1) / 2){
                flag = 1;
                break;
            }
            if(res <= (r[i] - l[i] + 1) / 2)update(1, 1, n, 0, l[i], r[i]);
            else update(1, 1, n, 1, l[i], r[i]);
        }
        if(flag) cout << "NO
";
        else {
            int sl = 1, sr = 1;
            while(sr <= n){
                while(s[sl] == s[sr]) sr++;
                int res = query(1, 1, n, sl, sr - 1);
                if(s[sl] == '0' && res != 0){
                    flag = 1;
                    break;
                }
                if(s[sl] == '1' && res != sr - sl){
                    flag = 1;
                    break;
                }
                sl = sr;
            }
            if(flag) cout << "NO
";
            else cout<<"YES
";
        }
    }
    return 0;
}

F. Nezzar and Nice Beatmap

题意：

(n)个点，重新排序使得任意三个连续的点形成的角（以中间的点为顶点）小于(90)度

思路：

任意选择一个点为起点，每次选择离该点最远的点作为下一个点，以此类推。假设第一个点为(a)，第二个点为(b)，则剩余的所有点在以(a)为圆心，(b)为半径的圆内，且(b)在圆上，则与下一个点形成的角一定小于(90)度

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 5010;
int vis[maxn], x[maxn], y[maxn];
ll dis2(int a, int b) {
    return (ll)(x[a] - x[b]) * (x[a] - x[b]) + (ll)(y[a] - y[b]) * (y[a] - y[b]);
}
vector<int> ans;
int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) cin >> x[i] >> y[i];
    int now = 0, m = n - 1;
    ans.push_back(now);
    vis[now] = 1;
    while (m--) {
        ll len = 0;
        int k = -1;
        for (int j = 0; j < n; j++) {
            if (vis[j] == 0 && dis2(now, j) > len) {
                len = dis2(now, j);
                k = j;
            }
        }
        vis[k] = 1;
        now = k;
        ans.push_back(now);
    }
    for (auto i : ans) printf("%d ", i + 1);
    puts("");
    return 0;
}