这道题要用到差分，因为反复经过一条路时只需要买一张对应的卡就行了，不用买多张，所以我们可以用差分，算出经过每条路的次数，要注意假设从1到3城市，只经过了道路1和道路2，应该让cha【1】++，cha【3】--；
还有算结果时应该从1到n-1列举每一条路，我最开始就搞错了，还要注意的一点就是两座城市的大小，差分是小的++，大的--；
#include<iostream>
#include<set>
#include<map>
#include<algorithm>
#include<vector>
#include<cmath>
#include<climits>
#include<cstring>
#define int long long
const int N = 1e6;
using namespace std;
char* p1, * p2, buf[100000];
#define nc() (p1==p2 && (p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
int read()
{int x = 0, f = 1;char ch = nc();while (ch < 48 || ch>57){if (ch == '-')f = -1;ch = nc();}while (ch >= 48 && ch <= 57)x = x * 10 + ch - 48, ch = nc();return x * f;
}
int a[N], b[N], c[N];
int suma[N], sumb[N], sumc[N];
int p[N];
int cha[N];
signed main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int n, m;cin >> n >> m;for (int i = 1; i <= m; i++)cin >> p[i];for (int i = 1; i <= n - 1; i++) {cin >> a[i] >> b[i] >> c[i];suma[i] = suma[i - 1] + a[i];sumb[i] = sumb[i - 1] + b[i];sumc[i] = sumc[i - 1] + c[i];}for (int i = 1; i <= m-1; i++) {int maxx = max(p[i], p[i + 1]);int minn = min(p[i], p[i + 1]);cha[minn]++;cha[maxx - 1 + 1]--;}for (int i = 1; i <= n - 1; i++)cha[i] += cha[i - 1];int sum = 0;for (int i = 1; i <= n-1; i++) {sum += min(a[i] * cha[i], cha[i] * b[i] + c[i]);}cout << sum;return 0;
}
