2020 Multi-University Training Contest 2（待补
2020 Multi-University Training Contest 2
1001 Total Eclipse
-
思路：最后删除的是最大的点，则按从大到小排序，每次加入点，先判断其相邻点是否已加入，有就合并，每次的贡献是已加入点个数*相邻点权的差值
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 1e5 + 10;
int t, n, m;
int w[N], pos[N], fa[N];
bool vis[N];
ll tot, ans;
vector<int> g[N << 2];
inline int find(int x){
if (x == fa[x])
return x;
return fa[x] = find(fa[x]);
}
inline bool cmp(int a, int b){
return w[a] > w[b];
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t -- ){
tot = 0, ans = 0;
memset(vis, false, sizeof(vis));
cin >> n >> m;
for (int i = 1; i <= n; i ++ ){
cin >> w[i];
pos[i] = fa[i] = i;
g[i].clear();
}
sort(pos + 1, pos + n + 1, cmp);
pos[n + 1] = 0;
for (int i = 1; i <= m; i ++ ){
int u, v;
cin >> u >> v;
g[v].push_back(u);
g[u].push_back(v);
}
for (int i = 1; i <= n; i ++ ){
tot ++ ;
vis[pos[i]] = true;
for (auto v: g[pos[i]]){
if (!vis[v])
continue;
int v_ = find(v), u_ = find(pos[i]);
if (v_ != u_){
fa[u_] = v_;
tot -- ;
}
}
ans += tot * (w[pos[i]] - w[pos[i + 1]]);
}
cout << ans << "\n";
}
return 0;
}
1006 The Oculus
-
思路：用ull 暴力即可
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 2e6 + 10;
int t, n, x;
ull s1, s2, s3;
ull f[N];
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
f[1] = 1, f[2] = 2;
for (int i = 3; i < N; i ++ )
f[i] = f[i - 1] + f[i - 2];
cin >> t;
while (t -- ){
s1 = s2 = s3 = 0;
cin >> n;
for (int i = 1; i <= n; i ++ ){
cin >> x;
s1 += x * f[i];
}
cin >> n;
for (int i = 1; i <= n; i ++ ){
cin >> x;
s2 += x * f[i];
}
cin >> n;
for (int i = 1; i <= n; i ++ ){
cin >> x;
s3 += x * f[i];
}
for (int i = 1; i <= n; i ++ ){
if (s1 * s2 == s3 + f[i]){
cout << i << "\n";
break;
}
}
}
return 0;
}
1010 Lead of Wisdom
-
思路：暴力dfs即可
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 60;
int T, n, k, t, tot;
ll a, b, c, d, ans;
struct node{
ll a, b, c, d;
}now;
map<int, int> mp;
vector<node> vec[N];
inline void dfs(int t, ll a, ll b, ll c, ll d){
if (t > tot){
ans = max(ans, a * b * c * d);
return ;
}
for (auto v: vec[t])
dfs(t + 1, a + v.a, b + v.b, c + v.c, d + v.d);
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> T;
while (T -- ){
tot = 0, ans = 0;
mp.clear();
for (int i = 0; i < N; i ++ )
vec[i].clear();
cin >> n >> k;
for (int i = 1; i <= n; i ++ ){
cin >> t >> a >> b >> c >> d;
if (!mp[t])
mp[t] = ++ tot ;
vec[mp[t]].push_back({a, b, c, d});
}
dfs(1, 100, 100, 100, 100);
cout << ans << "\n";
}
return 0;
}
