indeed 4.22 第一次网测

1.第一题 没有看

2. 由于数据范围很小，所以每一层需要全排列，寻找最小的花费，然后所有层加起来就是最后的结果。

 #include<bits/stdc++.h>
 #define pb push_back
 typedef long long ll;
 using namespace std;
 typedef pair<int, int> pii;
 const int maxn = 1e3 + ;
 int n;
 int m;
 int a[][][];
 int p[];
 
 void solve() {
     cin >> n >> m;
     ll res = ;
     for (int j = ; j < n; j++) {
             cin >> a[][j][] >> a[][j][];
         }
     for (int i = ; i <= m; i++) {
         for (int j = ; j < n; j++) {
             cin >> a[i][j][] >> a[i][j][];
         }
         if(i == ) continue;
         for (int j = ; j < n; j++)
             p[j] = j;
         ll mr = INT_MAX;
         do {
             ll t = ;
             for (int j = ; j < n; j++) {
                 t += abs(a[i][p[j] ][] - a[i - ][j][] ) + abs(a[i][p[j] ][] - a[i - ][j][] );
             }
             mr = min(t, mr);
         } while(next_permutation(p, p + n));
         res += mr;
     }
     cout << res << endl;
 
 }
 
 int main() {
     freopen("test.in", "r", stdin);
     //freopen("test.out", "w", stdout);
     ios::sync_with_stdio();
     cin.tie(); cout.tie();
     solve();
     return ;
 }

3.数据范围也是很小，枚举每一条边，然后存在不存在，进行判断累加。边的个数6 * 5  / 2 = 15. 1 << 15 = 32000. 然后乘上判断连通的复杂度，结果也是很小。

判断连通可以用bfs或者dfs， dsu都是可以的。

 #include<bits/stdc++.h>
 #define pb push_back
 typedef long long ll;
 using namespace std;
 typedef pair<int, int> pii;
 const int maxn = 1e3 + ;
 int n, m;
 //set<pii> se;
 int f[];
 int a[][];
 int b[][];
 void init() {
     for (int i = ; i <= n; i++) f[i] = i;
 }
 int fd(int x) {
     if(x == f[x]) return x;
     return f[x] = fd(f[x]);
 }
 bool check() {
     int sz = n;
     for (int i = ; i <= n; i++) {
         for (int j = i + ; j <= n; j++) {
             if(a[i][j]) {
                 int t1 = fd(i);
                 int t2 = fd(j);
                 if(t1 != t2) {
                     sz--;
                     f[t1] = t2;
                 }
             }
         }
     }
     return sz == ;
 }
 ll res;
 
 void work(int x, int y) {
     //cout << x << " " << y << endl;
     if(x > n) {
         init();
         res += check();
         return;
     }
     if(y > n) {
         work(x + , x + );
         return;
     }
     work(x, y + );
     if(b[x][y]) return;
     a[x][y] = a[y][x] = ;
     work(x, y + );
     a[x][y] = a[y][x] = ;
 }
 void solve() {
     cin >> n >> m;
     int x, y;
     for (int i = ; i < m; i++) {
         cin >> x >> y;
         //if(x > y) swap(x, y);
         //se.insert({x, y});
         b[x][y] = b[y][x] = ;
     }
     work(, );
     cout << res << endl;
 }
 
 int main() {
     freopen("test.in", "r", stdin);
     //freopen("test.out", "w", stdout);
     ios::sync_with_stdio();
     cin.tie(); cout.tie();
     solve();
     return ;
 }

4. 第四题稍微花了一些时间。 首先题目求解满足条件的最小， 那就是比这个大的都是可以满足条件的， 满足二分的性质。

然后给定答案，如何快速的判断是否满足要求， 贪心策略进行。由于 n = 1e4， 这个数其实挺小的， 排序， 贪心 可以的。

 #include<bits/stdc++.h>
 #define pb push_back
 typedef long long ll;
 using namespace std;
 typedef pair<int, int> pii;
 const int maxn = 1e4 + ;
 int n, a, b;
 ll h[maxn];
 ll tmp[maxn];
 int tot;
 bool check(ll x) {
     ll ra, rb;
     ra = rb = ;
     tot = ;
     for (int i = ; i < n; i++) {
         if(h[i] >= x) {
             ll c = h[i] / x;
             if(ra + c > a) {
                 ll t = h[i] - x * (a - ra);
                 ra = a;
                 if(t > ) tmp[tot++] = t;
             } else {
                 ra += h[i] / x;
                 int t = h[i] % x;
                 if(t > ) tmp[tot++] = t;
             }
 
         } else {
             tmp[tot++] = h[i];
         }
         //if(ra > a) return 0;
     }
     sort(tmp, tmp + tot, greater<ll>());
     for (int i = ; i < tot; i++) {
         if(ra < a) {
             ra++;
         } else {
             rb += tmp[i];
         }
     }
     return ra + rb <= b;
 }
 void solve() {
     cin >> n >> a >> b;
     for (int i = ; i < n; i++) {
         cin >> h[i];
     }
     sort(h, h + n, greater<ll>());
     ll left = , right = 1e9 + ;
     while(left < right) {
         ll mid = (left + right) / ;
         //cout << mid << " " << left << " " << right << endl;
         if(check(mid)) right = mid;
         else left = mid + ;
 
     }
     if(check(left)) cout << left << endl;
     else cout << - << endl;
 }
 
 int main() {
     //freopen("test.in", "r", stdin);
     //freopen("test.out", "w", stdout);
     ios::sync_with_stdio();
     cin.tie(); cout.tie();
     solve();
     return ;
 }