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;
}
}