L2-036 网红点打卡攻略

起初是忽略了每个地点只能打卡一次的条件。
错误版本:

#include <bits/stdc++.h>
using namespace std;
int edges[210][210], fangan[2000][2000];
int minspend = INT_MAX;
int idx = 0;
int main() {
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		edges[a][b] = edges[b][a] = c;
	}
	int t;
	cin >> t;
	for (int i = 1; i <= t; i++) {//第i种方案
		cin >> fangan[i][0];
		for (int j = 1; j <= fangan[i][0]; j++) {
			cin >> fangan[i][j];
		}
	}
	int res = 0;//可行方案
	for (int i = 1; i <= t; i++) {
		int spend = 0;
		if (fangan[i][0] != n) continue;
		//检查方案
		//是否能从家出发最后返回
		if (edges[0][fangan[i][1]] == 0 || edges[0][fangan[i][6]] == 0) continue;
		spend += edges[0][fangan[i][1]];
		spend += edges[0][fangan[i][6]];
		int j = 2;
		for (j = 2; j <= 6; j++) {
			if (edges[fangan[i][j]][fangan[i][j - 1]] == 0) {
				break;//不可达
			}
			spend += edges[fangan[i][j]][fangan[i][j - 1]];
		}
		if (j > fangan[i][0]) {
			res++;
			if (spend < minspend) {
				minspend = spend;
				idx = i;
			}
		}
	}
	cout << res << '\n';
	cout << idx << " " << minspend << '\n';
	return 0;
}

正确版本:

#include <bits/stdc++.h>
using namespace std;
int edges[210][210], s[210],vis[210];//s当前路线 vis地点是否已经来过
int minspend = INT_MAX;
int idx = 0;
int main() {
	int n, m;
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		edges[a][b] = edges[b][a] = c;
	}
	int count = 0;
	int t;
	cin >> t;
	for (int i = 1; i <= t; i++) {//第i种方案
		memset(vis, 0, sizeof(vis));
		int cnt;
		cin >> cnt;
		for (int j = 1; j <= cnt; j++) {
			cin >> s[j];
		}
		if (cnt != n) continue;
		int spend = 0;//旅行线路的花销
		int flag = 1;
		int u = 0;//u是家的位置
		for (int j = 1; j <= cnt; j++) {//检测i方案的可行性
			//如果已经来过或者不可达
			int v = s[j];
			if (vis[v] || edges[u][v] == 0) {//也就是避免一个点
				flag = 0;
				break;
			}
			vis[v] = 1;//设置该地点已经访问过
			spend += edges[u][v];
			u = v;
		}
		//是否能够到家
		if (edges[u][0] == 0) flag=0;
		if (flag==0) {//如果方案不可行
			continue;
		}
		count++;//方案可行 将方案数量增加
		spend += edges[u][0];
		if (spend < minspend) {//更新索引																																																																																																																																																																																																																																																																																																																																																																																																																																																																																																																																			       
			minspend = spend;
			idx = i;
		}
	}
	cout << count << '\n';
	cout << idx << " " << minspend << '\n';
	return 0;
}