2021牛客暑期多校训练营3

Posted 2021-08-27 TURNINING

B Black and white
思路：通过观察我们发现，我们其实最多只用选n + m - 1个点就好了，其它的点对答案的贡献为0。那么我们不是贪心的选择这n+m-1个格子就好了？还有一个问题，我们不能让一个小正方形4个点全都被选。这里有这样一个技巧，把一个点(i, j) 拆分成两个不同的点Ai Bj，当我们选择把这个点涂黑时就给Ai，Bj连一条边。只有正方形4个点全选才会改变连通性(可以自己画一下)，那么我们跑一次最小生成树就行了。

#include<bits/stdc++.h>
using namespace std;
 
#define lsn (A << 1)
#define rsn (A << 1 | 1)
#define mid (l + r >> 1)
 
typedef long long ll;
typedef unsigned long long ull;
 
typedef pair<int, int> P;
 
const int MAXN = 100000 + 10;
const int MAX_LEN = 100000 + 10;
const int MAX_LOG_V = 22;
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-7;
const ull B = 100000007;


int n, m, a, b, c, d, p;
int pre[MAXN];
vector<P> f[MAXN];

int find(int x) {
    if(pre[x] == 0) return x;
    return pre[x] = find(pre[x]);
}


void solve() {
    scanf("%d %d %d %d %d %d %d", &n, &m, &a, &b, &c, &d, &p);
    int last = a;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            int w = ((ll)last*last*b+(ll)last*c+d)%p;
            last = w;
            f[last].emplace_back(P(i,j+n));
        }
    }
    int ans = 0;
    for(int k = 0; k < p; k++) {
        for(int i = 0; i < f[k].size(); i++) {
            int u = f[k][i].first, v = f[k][i].second;
            int f1 = find(u), f2 = find(v);
            if(f1 != f2) {ans += k; pre[f1] = f2;}
        }
    }
    printf("%d\\n", ans);
}

int main() {
    //ios::sync_with_stdio(false);
    int t = 1; //scanf("%d", &t);
    while(t--) {
        solve();
    }
}

J Counting Triangles

思路：我们发现找不满足要求的比找满足要求的要好找的多，我们只要选一个边为1，另外一条为0，第三条边不管怎样都是不行的。每个不满足的三角形会被选两次((1,0)一次，(0,1)一次)所以要除以2

#include<bits/stdc++.h>
using namespace std;
 
#define lsn (A << 1)
#define rsn (A << 1 | 1)
#define mid (l + r >> 1)
 
typedef long long ll;
typedef unsigned long long ull;
 
typedef pair<int, int> P;
 
const int MAXN = 2000 + 10;
const int MAX_LEN = 100000 + 10;
const int MAX_LOG_V = 22;
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-7;
const ull B = 100000007;



namespace GenHelper
{
    unsigned z1,z2,z3,z4,b,u;
    unsigned get()
    {
        b=((z1<<6)^z1)>>13;
        z1=((z1&4294967294U)<<18)^b;
        b=((z2<<2)^z2)>>27;
        z2=((z2&4294967288U)<<2)^b;
        b=((z3<<13)^z3)>>21;
        z3=((z3&4294967280U)<<7)^b;
        b=((z4<<3)^z4)>>12;
        z4=((z4&4294967168U)<<13)^b;
        return (z1^z2^z3^z4);
    }
    bool read() {
      while (!u) u = get();
      bool res = u & 1;
      u >>= 1; return res;
    }
    void srand(int x)
    {
        z1=x;
        z2=(~x)^0x233333333U;
        z3=x^0x1234598766U;
        z4=(~x)+51;
      	u = 0;
    }
}
using namespace GenHelper;
bool edge[8005][8005];
int main() {
	int n, seed;
	cin >> n >> seed;
	srand(seed);
	for(int i = 0; i < n; i++)
		for(int j = i + 1; j < n; j++)
			edge[j][i] = edge[i][j] = read();
	ll ans = 0;
	for(int i = 0; i < n; i++) {
		int cnt1 = 0, cnt0 = 0;
		for(int j = 0; j < n; j++) {
			if(i == j) continue;
			if(edge[i][j]) cnt1++;
			else cnt0++;
		}
		ans += cnt0 * cnt1;
	}
	ans = (ll)n*(n-1)*(n-2)/6 - ans/2;
	cout << ans << "\\n";
 	return 0;
}