比赛链接：https://codeforces.com/contest/1559

总体评价：Chinese Round，前四道题目较为简单，后两道貌似难度剧增......

题意基本上是自己写的，也算练练英语吧......如果有语法错误还请海涵。

A

题意

Select an arbitrary interval \([l,r]\)and for all values \(i(0 \leq i \leq r - l)\), replace \(a_{l+i}\) with \(a_{l+i}\)&\(a_{r−i}\) at the same time.This operation can be performed any number of times. Minimize the maximum value in the sequence.

思路

将一个数用二进制表示后，每一位非\(1\)即\(0\)。

根据与运算的性质，两个数相与，对于二进制下都为\(1\)的某一位没有影响。然而如果存在第\(k\)位，一个数为\(1\)，另一个为\(0\)，那么结果必然会变小。

由于题目未限制与运算的次数，那么只需要让\(ans\)赋上\(a_1\)的值，让\(ans\)依次与之后的每一个数相与即可得到答案。

#include <cstdio>
#include <iostream>
const int N = 150;
using namespace std;
int t, n, ans, a[N];
int main() {
    ios::sync_with_stdio(false);
    cin >> t;
    while(t--) {
        cin >> n;
        for(int i = 1; i <= n; i++) cin >> a[i];
        ans = a[1];
        for(int i = 2; i <= n; i++) ans = ans & a[i];
        cout << ans << endl;
    }
    return 0;
}

B

题意

Given a string containing 'B', 'R', or '?'，you need to replace the '?' with 'B' or 'R' to get a new string.

At the same time, you should minimize the number of the ”BB“ and "RR".

思路

找到每个有字母的位置，利用贪心思想向前依次填字母。

需要特殊处理只有一个字母的情况，否则该字母后的"?"无法填上字母。

#include <bits/stdc++.h>
const int INF = 1 << 30;
const int N = 150;
typedef long long ll;
using namespace std;
struct node {
	int x, y;
}b[N];
int t, n, cnt, num, a[N];
char c;
int main() {
	cin >> t;
	while(t--) {
		num = cnt = 0;
		cin >> n;
		for(int i = 1; i <= n; i++) {
			cin >> c;
			if(c == '?') {
				a[i] = -1;
				++cnt;
			}
			else {
				a[i] = (c == 'B') ? 1 : 2;
				b[++num] = (node){i, cnt};
				cnt = 0;
			}
		}
		for(int i = 1; i <= num; i++) {
			for(int j = b[i].x - 1; j >= b[i].x - b[i].y; j--) {
				a[j] = (a[j + 1] == 1) ? 2 : 1;
			}
		}
		for(int i = 1; i <= n; i++) {
			if(a[i] == 1 || a[i] == 2)
				a[i] == 1 ? cout << "B" : cout << "R" ;
			else
				a[i - 1] == 1 ? (a[i] = 2, cout << "R") : (a[i] = 1, cout << "B");
		}
		cout << endl;
	}
	return 0;
}

C

题意

Follow the rules to add edges, then find a way which goes through every point exactly once.

思路

简简单单的建图、\(dfs\)。

或者也可以找规律来完成此题。

#include <bits/stdc++.h>
const int N = 10050;
using namespace std;
int t, n, num, a[N], ans[N];
bool flag, vis[N];
vector<int> g[N];
void dfs(int x, int step) {
	if(step == n && num - 1 == step) {
		flag = 1;
		for(int i = 1; i <= num; i++)
			cout << ans[i] << " ";
		cout << endl;
		return ;
	}
	for(int i = 0; i < g[x].size() && !flag; i++) {
		int to = g[x][i];
		if(vis[to])
			continue;
		vis[to] = 1;
		ans[++num] = to;
		dfs(to, step + 1);
		num--;
		vis[to] = 0;
	}
}
int main() {
	cin >> t;
	while(t--) {
		cin >> n;
		for(int i = 1; i <= n - 1; i++)
			g[i].push_back(i + 1);
		for(int i = 1; i <= n; i++) {
			cin >> a[i];
			a[i] ? g[n + 1].push_back(i) : g[i].push_back(n + 1); 
		}
		for(int i = 1; i <= n + 1; i++) {
			if(flag)
				break;
			memset(vis, 0, sizeof(vis));
			vis[i] = 1;
			num = 1;
			ans[num] = i;
			dfs(i, 0);
		}
		if(!flag)
			cout << "-1" << endl;
		for(int i = 1; i <= n + 1; i++)
			g[i].clear();
		memset(vis, 0, sizeof(vis));
		flag = 0;
	}
	return 0;
}

D1

题意

Given two forests(A forest is an undirected graph without cycles (not necessarily connected).), you can add an edge between \(u\) and \(v\)\((u, v \in [1, n])\) in two forests at the same time.Maximum the number of edges they can add, and which edges to add.

思路

考虑到数据范围，一种很暴力的思路就是邻接矩阵存图，\(O(n^2)\)枚举边，每次\(dfs\)判环，然后愉快的Time limit exceeded on pretest 6。

正确做法是不用建图，只需要使用并查集即可，加上路径压缩的并查集可以节省很多时间。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
const int N = 1050;
using namespace std;
int n, m1, m2, u, v, fa[2][N];
struct node {
    int x, y;
};
vector<node> ans;
int get_fa(bool tp, int x) { return x == fa[tp][x] ? x : fa[tp][x] = get_fa(tp, fa[tp][x]); }
int main() {
	ios::sync_with_stdio(false);
    cin >> n >> m1 >> m2;
    for(int i = 1; i <= n; i++) fa[0][i] = fa[1][i] = i;
    for(int i = 1; i <= m1; i++) {
        cin >> u >> v;
        fa[0][get_fa(0, v)] = get_fa(0, u);
    }	
    for(int i = 1; i <= m2; i++) {
        cin >> u >> v;
        fa[1][get_fa(1, v)] = get_fa(1, u);
    }
    for(int i = 1; i <= n; i++) {
        for(int j = i + 1; j <= n; j++) {
            if((get_fa(0, i) != get_fa(0, j)) && (get_fa(1, i) != get_fa(1, j))) {
                fa[0][get_fa(0, j)] = get_fa(0, i);
                fa[1][get_fa(1, j)] = get_fa(1, i);
                ans.push_back((node){i, j});
            }
        }
    }
    cout << ans.size() << endl;
    for(int i = 0; i < ans.size(); i++)
        cout << ans[i].x << " " << ans[i].y << endl;
    return 0;
}