Topic
C:
Violent search
#include<bits/stdc++.h>
using namespace std;
int n, k, r, ans;
struct ty
{
int x, y, r;
}a[20], pos[10];
bool xiangjiao(int i, int j)
{
// 两圆相切 可以用勾股定理判断
if ((a[i].x-pos[j].x)*(a[i].x-pos[j].x)+(a[i].y-pos[j].y)*(a[i].y-pos[j].y) <= (a[i].r+r)*(a[i].r+r))
return 1;
return 0;
}
int calc()
{
int cnt = 0;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= k; j++)
{
if (xiangjiao(i, j))
{
cnt++;
break;
// 此时存了 k 个点，一但某个所给的点可以被k个点中某个打到，就break
// 保证了一个点不会被重复计数
}
}
return cnt;
}
void dfs(int dep)
{
if (dep > k)// 存好所选的k个点，去判断这k个点最多消灭多少敌人
{
ans = max(ans, calc());
return ;
}
for (int i = -7; i <= 7; i++)
for (int j = -7; j <= 7; j++)
{
pos[dep].x = i;
pos[dep].y = j;
dfs(dep + 1);
}
}
int main()
{
scanf("%d %d %d", &n, &k, &r);
for (int i = 1; i <= n; i++)
scanf("%d %d %d", &a[i].x, &a[i].y, &a[i].r);
dfs(1);
cout << ans << endl;
return 0;
}
G question
and check set
#include<iostream>
#include<cstdio>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int Max = (int)1e5 + 9;
int t,n,m;
//int bcj[Max];
struct xpy
{
ll a;
int x;
}bcj[1000009];
int Find(int x)
{
int r = x, j, k = x;
while(r != bcj[r].x) r = bcj[r].x; // 寻找根节点
while(k != r)
{
j = bcj[k].x; // 暂存此时k的父节点
bcj[k].x = r; // 将k的父节点改为根节点
bcj[r].a = max(bcj[k].a,bcj[r].a);
k = j; // k移到父节点 直至全改为根节点
}
return r;
}
void Union(int x,int y)
{
x = Find(x);// x变成x的根节点
y = Find(y);// y变成y的根节点
if(x != y) bcj[x].x = y;
}
void work()
{
for(int i = 1; i <= n; i++)
{
scanf("%lld", &bcj[i].a);
bcj[i].x = i;
}
int x, y, t;
while(m--)
{
scanf("%d %d", &x, &y);
Union(x, y);
}
for(int i = 1; i <= n; i++) Find(i);// 更新根节点 保证这个集合的最大值更新到根节点身上
ll sum = 0;
for(int i = 1; i <= n; i++)
{
sum += bcj[Find(i)].a;
//printf("%d ", Find(i));
}
cout << sum << endl;
}
int main()
{
scanf("%d %d", &n, &m);
work();
return 0;
}
/*
6 4
1 2 3 4 5 6
1 2
2 3
4 5
5 6
*/
I problem
method one
dynamic programming
#include<bits/stdc++.h>
using namespace std;
const int N= 1e6+7;
int n,a[N],last[N], f[N];
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
{
scanf("%d", &a[i]);
// 用数组last存某一个数字上一次出现的位置
if(last[a[i]] != 0) f[i] = max(f[i-1], f[last[a[i]]]+1);
else f[i] = f[i-1];
last[a[i]] = i;
}
cout << f[n];
return 0;
}
Method Two
Greedy
#include <bits/stdc++.h>
using namespace std;
int a[1000010];//
map<int, int> m;
int main()
{
int n;
cin >> n;
for (int i = 1; i <= n; i++)
cin >> a[i];
long long sum = 0;
for (int i = 1; i <= n; i++)
{
// cout<<i<<" "<<a[i]<<" "<<m[a[i]]<< endl;
if (m[a[i]])
{
sum++;
m.clear();
}
m[a[i]]=1;
}
cout<<sum<<endl;
return 0;
}
Question J:
Game
Move forward from the final state and find some rules
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const int Max = 1000009;
int n;
int sum1,sum2;
ll a[Max];
int main()
{
int b;
cin >> n;
int sum = 0;
for(int i = 1; i <= n; ++i)
{
int q;
scanf("%d", &q);
if(q % 2 == 0) sum++;
}
if(n == 1 && sum == 0) cout << "NiuNiu\n";
else if(n == 1 && sum == 1) cout << "NiuMei\n";
else if(sum >= 2) cout << "NiuMei\n";
else
{
if(n % 2 != 0) cout << "NiuMei\n";
else cout << "NiuNiu\n";
}
return 0;
}