Codeforces Round #588 (Div. 2)
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C:
本题可以直接暴力求解,类比八皇后问题,使每个点分别对应一个值,分别判断有几条边能够满足即可
#include<bits/stdc++.h>
using namespace std;
#define ms(x,y) memset(x, y, sizeof(x))
#define lowbit(x) ((x)&(-x))
#define sqr(x) ((x)*(x))
typedef long long LL;
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
bool vis[8][8];
int col[8];
void run_case() {
int n, m;
cin >> n >> m;
vector<pii> a(m);
for(int i = 0; i < m; ++i)
cin >> a[i].first >> a[i].second;
int ans = 0;
function<void(int)> dfs = [&](int x) {
if(x > n) {
int cnt = 0;
ms(vis, 0);
for(int i = 0; i < m; ++i) {
if(vis[col[a[i].first]][col[a[i].second]]) continue;
vis[col[a[i].first]][col[a[i].second]] = vis[col[a[i].second]][col[a[i].first]] = true;
cnt++;
}
ans = max(ans, cnt);
return;
}
for(int i = 1; i <= 6; ++i) {
col[x] = i;
dfs(x+1);
}
};
dfs(1);
cout << ans;
}
int main() {
ios::sync_with_stdio(false), cin.tie(0);
cout.flags(ios::fixed);cout.precision(9);
//int t; cin >> t;
//while(t--)
run_case();
cout.flush();
return 0;
}
D:
读懂题意后也不难,至少有2个人是calm的,则集合中至少有2个人的knowledge是相等的,设为(x),然后再遍历,设(y)为当前的值,若(x&y==y),则说明(x geq y)且(y)中拥有的knowledge(x)都有
#include<bits/stdc++.h>
using namespace std;
#define ms(x,y) memset(x, y, sizeof(x))
#define lowbit(x) ((x)&(-x))
#define sqr(x) ((x)*(x))
typedef long long LL;
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
void run_case() {
int n; cin >> n;
vector<LL> a(n), b(n), valid;
map<LL, int> mp;
for(auto &x: a) {
cin >> x;
mp[x]++;
}
for(auto &x: b) cin >> x;
for(auto i: mp)
if(i.second > 1)
valid.push_back(i.first);
LL ans = 0;
for(int i = 0; i < n; ++i) {
for(auto j: valid) {
if((a[i]&j)==a[i]) {
ans += b[i];
break;
}
}
}
cout << ans;
}
int main() {
ios::sync_with_stdio(false), cin.tie(0);
cout.flags(ios::fixed);cout.precision(9);
//int t; cin >> t;
//while(t--)
run_case();
cout.flush();
return 0;
}
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