两日总结十七

前面放了几天假，这两天主要就是补了一下子前面比赛的题目。

首先是牛客蔚蓝杯最后一场的题目：

 这一场是真的太可惜了，真的可惜，有两个题目都是直接想出来的，但是就是想的有的复杂。

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Yet Another FFT Problem?

这个一开始直接推出来就是找ai + bj == ai2 + bj2就行了，被题目影响了没有直接去FFT了，然后2e7直接爆内存了。。。其实就是暴力就行了，因为鸽巢原理，最多不会超过2e7+1个，找到就退出就行了，当时SB了。

#include 
using namespace std;
const int N = 1e6 + 5, inf = 0x3f3f3f3f, M = 1e7 + 5;
 
int A[N],B[N];
vectorint,int>> a,b,ans;
int n,m,x;
int mp1[2*M], mp2[M];
 
int main() {
    cin >> n >> m;
    int fa = -1, fb = -1;
    for(int i =  1; i <= n; i ++) {
        cin >> x;
        A[i] = x;
        if(!mp1[x]) mp1[x] = i, a.push_back({x,i});
        else if(fa == -1) fa = i;
    }
    for(int i = 1; i <= m; i ++) {
        cin >> x;
        B[i] = x;
        if(!mp2[x]) mp2[x] = i, b.push_back({x,i});
        else if(fb == -1) fb = i;
    }
    if(fa != -1 && fb != -1) {
        cout << mp1[A[fa]] << " " << fa << " " << mp2[B[fb]] << " " << fb << endl;
    }else {
        for(int i = 0; i <= M; i ++) mp1[i]= 0;
        int cnt = 1;
        for(auto x : a) {
            for(auto y : b) {
                if(mp1[x.first + y.first]) {
                    cout << ans[mp1[x.first+y.first]-1].first << " " << x.second << " " << ans[mp1[x.first+y.first]-1].second << " " << y.second << endl;
                    return 0; 
                }
                mp1[x.first + y.first] = cnt;
                ans.push_back({x.second,y.second});
                cnt ++;
            }
        }
        cout << -1 << endl;
    }
    return 0;
}

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然后就是这个Shannon Switching Game?也是相当怎么搞，但是细节没有处理好，不应该只往一边涂色的。。。

#include 
 
#define OST  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#define ll long long int
#define ull unsigned long long int
using namespace std;
int e[105][105];
int vis[105];
int n, m, s, t;
 
void BFS() {
	queue<int> q;
	q.push(t);
	while (q.size()) {
		int now = q.front();
		q.pop();
		for (int i = 1; i <= n; i ++) {
			if (vis[i] == t) continue;
			if (vis[i] != t && e[now][i] >= 2) {
				vis[i] = t;
				q.push(i);
			}
			int cnt = 0;
			for (int j = 1; j <= n; j ++) {
				if (vis[j] == t && e[i][j]) {
					cnt ++;
				}
			}
			if (cnt >= 2) vis[i] = t, q.push(i);
		}
	}
}
 
void BFS2() {
	queue<int> q;
	q.push(s);
	while (q.size()) {
		int now = q.front();
		q.pop();
		for (int i = 1; i <= n; i ++) {
			if (vis[i] != s && e[now][i] >= 2) {
				vis[i] = s;
				q.push(i);
			}
		}
	}
}
 
void solve() {
 
	cin >> n >> m >> s >> t;
	for (int i = 1; i <= n; i ++) vis[i] = i;
	for (int i = 1; i <= n; i ++)
		for (int j  = 1; j <= n; j ++) e[i][j] = 0;
	for (int i = 1; i <= m; i ++) {
		int x, y;
		cin >> x >> y;
		e[x][y] ++;
		e[y][x] ++;
	}
	BFS();
	// for (int i = 1; i <= n; i ++) {
	// 	cout << vis[i] << " ";
	// }
	if (vis[s] == t) {
		cout << "Join Player" << endl;
	} else {
		BFS2();
 
		for (int i = 1; i <= n; i ++) {
			if (vis[i] == s) {
				int cnt = 0;
				for (int i = 1; i <= n; i++) {
					if (vis[i] == t && e[s][i]) {
						cnt++;
					}
				}
				if (cnt >= 2) {
					cout << "Join Player" << endl;
					return;
				}
			}
		}
		cout << "Cut Player" << endl;
	}
}
int main(int argc, char const *argv[]) {
	OST;
	int T;
	cin >> T;
	while (T --) {
		solve();
	}
	return 0;
}

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还有这个Reviewer Assignment网络流的板子题目，主要是建立边有点复杂，遇到的也少，就压根没有想到。最大流还有最小费用流都可以写。

然后把这个题目补了，顺便重新整理了一下最大流，还有费用流的模板。

 然后就是Codeforces Round #816 (Div. 2)的补题：

E题真的巧妙，每一次改变之后跑一遍最短路，然后每次更新最短值，最后面更新K次就行了。

#include 
 
 
#define OST  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#define ll long long int
#define ull unsigned long long int
#define pair pair
using namespace std;
const int N = 1e5 + 10;
const ll inf = 0x3f3f3f3f3f3f3f3f;
 
int n, m, k;
struct node {
	int y;
	ll v;
};
 
vector e[N];
vectordis(N, inf), dis2(N, inf);
 
 
void Dij() {
	priority_queue, greater> q;
	for (int i = 1; i <= n; i ++) q.push({dis[i], i});
	while (q.size()) {
		auto [v, now] = q.top();
		q.pop();
		for (auto x : e[now]) {
			if (dis[x.y] > v + x.v) {
				dis[x.y] = v + x.v;
				q.push({dis[x.y], x.y});
			}
		}
	}
}
 
void Fen(int l, int r, int L, int R) {
	// cout << l << " " << r << " " << L << " " << R << endl;
	if (l > r) return;
	int mid = (l + r) / 2;
	//找到传送一次到m最小的
	int idx = -1;
	ll maxn = inf;
	for (int i = L; i <= R; i ++) {
		ll tmp = dis[i] + 1ll * pow(i - mid, 2);
		if (maxn > tmp) {
			maxn = tmp;
			idx = i;
		}
	}
	dis2[mid] = maxn;
	Fen(l, mid - 1, L, idx);
	Fen(mid + 1, r, idx, R);
}
 
 
int main(int argc, char const *argv[]) {
	OST;
	cin >> n >> m >> k;
	for (int i = 1; i <= m; i ++) {
		int x, y, v;
		cin >> x >> y >> v;
		e[x].push_back({y, v});
		e[y].push_back({x, v});
 
	}
	dis[1] = 0;
	Dij();
	// for (int  i = 1; i <= n; i ++) cout << dis[i] << " ";
	// cout << endl;
	for (int i = 1; i <= k; i ++) {
		for (int i = 1; i <= n; i ++) dis2[i] =  inf;
		Fen(1, n, 1, n);
		for (int i = 1; i <= n; i ++) dis[i] = dis2[i];
		Dij();
	}
	for (int  i = 1; i <= n; i ++) cout << dis[i] << " ";
	cout << endl;
	return 0;
}

 然后就是这几天1700分的CF题目：