2020/3/19 刷题

最终刷题太多了博客思路慢慢更新

1.利用havel定理判断可不可以简单图化的一道好题

Sample Input

    3
    7
    4 3 1 5 4 2 1 
    6
    4 3 1 4 2 0 
    6
    2 3 1 1 2 1 

Sample Output

    YES
    0 1 0 1 1 0 1 
    1 0 0 1 1 0 0 
    0 0 0 1 0 0 0 
    1 1 1 0 1 1 0 
    1 1 0 1 0 1 0 
    0 0 0 1 1 0 0 
    1 0 0 0 0 0 0 

    NO

    YES
    0 1 0 0 1 0 
    1 0 0 1 1 0 
    0 0 0 0 0 1 
    0 1 0 0 0 0 
    1 1 0 0 0 0 
    0 0 1 0 0 0 

#include<iostream>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
bool MAP[15][15];
typedef pair<int, int> P;
std::vector<P> v;
bool cmp1(const P& p1, const P& p2) {
	return p1.second > p2.second;
}
int main (){
	int n;
	cin >> n;
	while(n--) {
		v.clear();
		memset(MAP, 0, sizeof(MAP));
		int m;
		cin >> m;
		for(int i = 1; i <= m; ++i) {
			int temp;
			cin >> temp;
			v.push_back(P(i, temp));
		}
		//sort(v.begin(), v.end(), cmp1);
		// for(int i = 0; i < m; ++i) 
		// {
		// 	cout << v[i].second << " "; 
		// }
		bool flag = 0;
		for(int i = 0; i < m; ++i) {
			sort(v.begin() + i, v.end(), cmp1);
			for(int j = 0; j < v[i].second; ++j) {
				if(i + 1 + j < m){
					v[i + 1 + j].second--;
					if(v[i+1+j].second < 0){
						flag = 1;
						break;
					}
					MAP[v[i].first][v[i+1+j].first] = 1;
					MAP[v[i+1+j].first][v[i].first] = 1;
				}
				else {
					flag = 1;
					break;
				}
			}
			if(flag) {
				break;
			}
		}
		if(v[m - 1].second == 1) {
			flag = 1;
		}
		if(flag) {
			cout << "NO" << endl;
		}
		else {
			cout << "YES" << endl;
			for(int i = 1; i <= m; ++i) {
				for(int j = 1; j <= m; ++j) {
					cout << MAP[i][j] << " ";
				}
				cout << endl;
			}
		}
		if(n != 0) {
			cout << endl;
		}
	}
}

 2.

Sample Input

    4
    1 2
    3 4
    5 6
    1 6
    4
    1 2
    3 4
    5 6
    7 8

Sample Output

    4
    2

//并查集裸题
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 10000005;
int ans;
int fa[MAXN];
int num[MAXN];
void init() {
	for(int i = 1; i <= 10000000; ++i) {
		fa[i] = i;
		num[i] = 1;
	}
}
int find(int x) {
	return x == fa[x] ? x : (fa[x] = find(fa[x]));
}
void merge(int i, int j) {
	int x = find(i);
	int y = find(j);
	if(x != y) {
		fa[x] = y;
		num[y] += num[x];	//合并十分关键
	}
}
int main () {
	int m;
	while(cin >> m) {
		if(!m){
			cout << 1 << endl;
			continue;
		}
		init();
		int mm = 0;
		while(m--) {
			int A, B;
			cin >> A >> B;
			mm = max(max(A,B), mm);
			merge(A, B);
		}
		int temp = 0;
		for(int i = 1; i <= mm; ++i) {
			temp = max(num[i], temp);
		}
		cout << temp << endl;
	}
}

 3.

Sample Input

    1 1
    1 1
    2 1
    1 2

Sample Output

    1
    26

//特别好的一道思维题。并查集加快速幂
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MODE = 1000000007;
const int MAXN = 1e7 + 5;
int fa[MAXN];
void init() {
	for(int i = 0; i <= 10000000; ++i) {
		fa[i] = i;
	}
}

ll find(ll x) {
	return x == fa[x] ? x : (fa[x] = find(fa[x]));
}

int merge(ll i, ll j) {
	ll x = find(i);
	ll y = find(j);
	if(x != y) {
		fa[x] = y;
		return 1;
	}
	return 0;
}
ll qpower(ll a, ll b) {
	ll r = 1;
	while(b) {
		if(b & 1) {
			r = (r * a) % MODE;
		}
		a = (a * a) % MODE;
		b >>= 1;
	}
	return r;
}

int main () {
	ll N, M;
	while(cin >> N >> M) {
		init();
		while(M--) {
			ll L, R;
			cin >> L >> R;
			N -= merge(L - 1, R);
		}
		cout << qpower(26, N) << endl;	
	}
}

 4.

//白给题
#include <bits/stdc++.h>
using namespace std;
multimap<int, int> e;
int main () {
	int n;
	while(cin >> n && n) {
		e.clear();
		for(int i = 0; i < n; ++i) {
			int f, t;
			cin >> f >> t;
			bool flag = true;
			for(multimap<int, int>::iterator i = e.find(t); i != e.end() && i -> first == t; ++i) {
				if(i -> second == f) {
					e.erase(i);
					flag = false;
					break;
				}
			}
			if(flag) {
				e.insert(make_pair(f, t));
			}
		}
		if(e.empty()) {
			cout << "YES" << endl;
		}
		else cout << "NO" << endl;
	}
}

 5.

#include <bits/stdc++.h>
using namespace std;
double dp[2][105];
int main () {
	int n, m, c;
	dp[0][0] = 1;
	while(cin >> n >> m >> c && (n + m)) {
		for(int i = 1; i <= c; ++i) {
//dp[0][C]代表总共有C个糖 按如上处理最后是偶数的概率
//dp[1][C]代表总共有C个糖 按如上处理最后是偶数的概率
/*
状态转移方程是 此时最后是偶数的概率等于 
原来是奇数的概率乘以这个糖落在n的部分概率即n/m+n 
加上 原来是偶数的概率乘以这个糖落在n的部分概率即m/m+n
*/
			dp[0][i] = dp[1][i-1] * (n/(n+m+0.0)) + 
			dp[0][i-1]*(m/(n+m+0.0));

			dp[1][i] = dp[0][i-1]*(n/(n+m+0.0)) +
			dp[1][i-1]*(m/(n+m+0.0));
		}
		cout << setprecision(8) << dp[0][c] << endl;
	}
}