31.P1636 Einstein学画画
此题为欧拉通路，必须要满足奇点的个数为0或2个
奇点：度数（入度出度）为奇数的点
如果奇点为2个或者0个就可以直接一笔化成
eg.
我们发现奇数点个数每增加2个就多一笔
#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, m, a, b, du[N], ans;
int main()
{ cin >> n >> m;for(int i = 1; i <= m; i ++){cin >> a >> b;du[a] ++, du[b] ++;}for(int i = 1; i <= n; i ++){if(du[i] & 1)ans ++;}cout << max(1, ans / 2);//最少也得画一笔 return 0;
}32. P1652 圆
我们枚举每一个圆，判断这个圆包含1和2哪个点，如果如果都包含说明这个线不会穿过这个圆，否则就将这个圆的穿线个数+1
#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, x[N], y[N], r[N], xx1, yy1, xx2, yy2, ans;
int dist(int xx, int yy, int xxx, int yyy)
{return (xxx - xx) * (xxx - xx) + (yyy - yy) * (yyy - yy);
}
int main()
{cin >> n;for(int i = 1; i <= n; i ++)cin >> x[i];for(int i = 1; i <= n; i ++)cin >> y[i];for(int i = 1; i <= n; i ++)cin >> r[i];cin >> xx1 >> yy1 >> xx2 >> yy2;for(int i = 1; i <= n; i ++){bool d1 = dist(xx1, yy1, x[i], y[i]) <= r[i] * r[i];bool d2 = dist(xx2, yy2, x[i], y[i]) <= r[i] * r[i];if(d1 ^ d2)ans ++;}cout << ans;return 0;
}33.P1657 选书
对书进行dfs，看是选a还是选b
#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int x, ans, a[N], b[N];
bool v[N];
void dfs(int dep)
{if(dep == x + 1){ans ++;return;}if(!v[a[dep]]){v[a[dep]] = 1;dfs(dep + 1);v[a[dep]] = 0;}if(!v[b[dep]]){v[b[dep]] = 1;dfs(dep + 1);v[b[dep]] = 0;}
}
int main()
{cin >> x;for(int i = 1; i <= x; i ++){cin >> a[i] >> b[i];}dfs(1);cout << ans << '\n';return 0;
}34.P1795 无穷的序列
找规律，将为1的数的下标先预处理出来
#include<bits/stdc++.h>
using namespace std;
vector<int> v;
int n, x;
int main()
{ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);for(int i = 1, j = 0; i <= 1e9; i += j ++)v.push_back(i);cin >> n;for(int i = 1; i <= n; i ++){cin >> x;auto it = lower_bound(v.begin(), v.end(), x);bool flag = (it != v.end() && *it == x);if(flag)cout << 1 << '\n';else cout << 0 << '\n';}return 0;
}35.P1757 通天之分组背包
我们可以将分出的组别用v[N]来记录，将同一组存入同一个v[N]中去，从而转化为背包问题
dp[i][j]表示到i组为止，总重量为j的最大价值
#include<bits/stdc++.h>
using namespace std;
const int N = 2e3 + 10;
vector<pair<int, int>> v[N];
int n, m, a, b, c, dp[N][N];
int main()
{cin >> m >> n;for(int i = 1; i <= n; i ++){cin >> a >> b >> c;v[c].push_back({a, b});}for(int i = 1; i <= n; i ++){for(int j = 0; j <= m; j ++){dp[i][j] = dp[i - 1][j];for(auto k : v[i])if(j >= k.first)dp[i][j] = max(dp[i][j], dp[i - 1][j - k.first] + k.second);}}cout << dp[n][m];return 0;
}
