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 + 10;
int n;
int m;
int a[110][10][2];
int p[10];

void solve() {
    cin >> n >> m;
    ll res = 0;
    for (int j = 0; j < n; j++) {
            cin >> a[0][j][0] >> a[0][j][1];
        }
    for (int i = 1; i <= m; i++) {
        for (int j = 0; j < n; j++) {
            cin >> a[i][j][0] >> a[i][j][1];
        }
        if(i == 0) continue;
        for (int j = 0; j < n; j++)
            p[j] = j;
        ll mr = INT_MAX;
        do {
            ll t = 0;
            for (int j = 0; j < n; j++) {
                t += abs(a[i][p[j] ][0] - a[i - 1][j][0] ) + abs(a[i][p[j] ][1] - a[i - 1][j][1] );
            }
            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(0);
    cin.tie(0); cout.tie(0);
    solve();
    return 0;
}

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 + 10;
int n, m;
//set<pii> se;
int f[10];
int a[10][10];
int b[10][10];
void init() {
    for (int i = 0; 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 = 1; i <= n; i++) {
        for (int j = i + 1; j <= n; j++) {
            if(a[i][j]) {
                int t1 = fd(i);
                int t2 = fd(j);
                if(t1 != t2) {
                    sz--;
                    f[t1] = t2;
                }
            }
        }
    }
    return sz == 1;
}
ll res;

void work(int x, int y) {
    //cout << x << " " << y << endl;
    if(x > n) {
        init();
        res += check();
        return;
    }
    if(y > n) {
        work(x + 1, x + 2);
        return;
    }
    work(x, y + 1);
    if(b[x][y]) return;
    a[x][y] = a[y][x] = 1;
    work(x, y + 1);
    a[x][y] = a[y][x] = 0;
}
void solve() {
    cin >> n >> m;
    int x, y;
    for (int i = 0; i < m; i++) {
        cin >> x >> y;
        //if(x > y) swap(x, y);
        //se.insert({x, y});
        b[x][y] = b[y][x] = 1;
    }
    work(1, 2);
    cout << res << endl;
}

int main() {
    freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    solve();
    return 0;
}

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 + 10;
int n, a, b;
ll h[maxn];
ll tmp[maxn];
int tot;
bool check(ll x) {
    ll ra, rb;
    ra = rb = 0;
    tot = 0;
    for (int i = 0; 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 > 0) tmp[tot++] = t;
            } else {
                ra += h[i] / x;
                int t = h[i] % x;
                if(t > 0) tmp[tot++] = t;
            }

        } else {
            tmp[tot++] = h[i];
        }
        //if(ra > a) return 0;
    }
    sort(tmp, tmp + tot, greater<ll>());
    for (int i = 0; i < tot; i++) {
        if(ra < a) {
            ra++;
        } else {
            rb += tmp[i];
        }
    }
    return ra + rb <= b;
}
void solve() {
    cin >> n >> a >> b;
    for (int i = 0; i < n; i++) {
        cin >> h[i];
    }
    sort(h, h + n, greater<ll>());
    ll left = 1, right = 1e9 + 10;
    while(left < right) {
        ll mid = (left + right) / 2;
        //cout << mid << " " << left << " " << right << endl;
        if(check(mid)) right = mid;
        else left = mid + 1;

    }
    if(check(left)) cout << left << endl;
    else cout << -1 << endl;
}

int main() {
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    solve();
    return 0;
}