C - Make Isomorphic
此题思路为，利用全排列枚举所有点的关系，如果需要调整则调整，不需要就继续，代码如下：

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  #include <bits/stdc++.h>
using ll = long long;
using namespace std;
const ll N = 1e6 + 5;

ll n, mg, mh;
bool g[10][10], h[10][10];
int res[10];
int a[10][10];
ll ans = 1e18;
int used[10];

void dfs(int step) {
	if (step > n) {
		ll cost = 0;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				if (g[res[i]][res[j]] != h[i][j]) {
					cost += a[i][j];
				}
			}
		}
		ans = min(ans, cost);
		return;
	}
	for (int i = 1; i <= n; i++) {
		if (!used[i]) {
			used[i] = 1;
			res[step] = i;
			dfs(step + 1);
			used[i] = 0;
		}
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cin >> n;
	cin >> mg;
	for (int i = 1; i <= mg; i++) {
		int u, v;
		cin >> u >> v;
		g[u][v] = 1;
		g[v][u] = 1;
	}
	cin >> mh;
	for (int i = 1; i <= mh; i++) {
		int u, v;
		cin >> u >> v;
		h[u][v] = 1;
		h[v][u] = 1;
	}
	for (int i = 1; i <= n - 1; i++) {
		for (int j = i + 1; j <= n; j++) {
			cin >> a[i][j];
		}
	}
	dfs(1);
	cout << ans << endl;
	return 0;
}

D - 1D Country
此题使用二分查找或lower_bound和upper_bound找到区间内的村庄个数，然受前缀和即可找出答案，代码如下：

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  #include <bits/stdc++.h>

using namespace std;
using ll = long long;
#define endl '\n'
const ll N = 2e5 + 5;
ll sum[N];
int main() {
	int n;
	cin >> n;
	int x[N], p[N];
	for (int i = 1; i <= n; i++) {
		cin >> x[i];
	}
	for (int i = 1; i <= n; i++) {
		cin >> p[i];
		sum[i] = sum[i - 1] + p[i];
	}
	int t;
	cin >> t;
	while(t--) {
		int pos1, pos2;
		int l, r;
		cin >> l >> r;
		pos1 = lower_bound(x + 1, x + n + 1, l) - x - 1;
		pos2 = upper_bound(x + 1, x + n + 1, r) - x - 1;
		ll ans = sum[pos2] - sum[pos1];
		cout << ans << endl;
	}
	return 0;
}

后面会补补后面的题，等更新吧