题目链接：传送门
思路：
仔细想一下可以发现：每个位置最多就增加2个高度。
所以就可以有状态：
f[i][j]表示保证前i个篱笆都是great时，第i个篱笆增加j的高度所需要的最小花费（1 <= i <= n, 0 <= j <= 2）。总共有3n个状态。
如果i = 1，f[i][j] = a[1] * j；
如果i > 1, f[i][j] = min{f[i-1][k] | 0 <= k <= 2 && a[i]+j != a[i-1]+k}；这里的“a[i]+j != a[i-1]+k”保证了篱笆都是great的。
答案ans = min{f[n][j] | 0 <= j <= 2}
时间复杂度是O(n)的。
代码：O(n)
#include <bits/stdc++.h> #define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0) #define N 300005 #define M 100005 #define INF 0x3f3f3f3f #define mk(x) (1<<x) // be conscious if mask x exceeds int #define sz(x) ((int)x.size()) #define lson(x) (x<<1) #define rson(x) (x<<1|1) #define mp(a,b) make_pair(a, b) #define endl ' ' #define lowbit(x) (x&-x) using namespace std; typedef long long ll; typedef double db; /** fast read **/ template <typename T> inline void read(T &x) { x = 0; T fg = 1; char ch = getchar(); while (!isdigit(ch)) { if (ch == '-') fg = -1; ch = getchar(); } while (isdigit(ch)) x = x*10+ch-'0', ch = getchar(); x = fg * x; } template <typename T, typename... Args> inline void read(T &x, Args &... args) { read(x), read(args...); } ll a[N], b[N]; ll f[N][3]; int main() { int q; cin >> q; while (q--) { int n; cin >> n; for (int i = 1; i <= n; i++) { read(a[i], b[i]); memset(f[i], 0x3f, sizeof(f[i])); } memset(f[0], 0, sizeof(f[0])); for (int i = 1; i <= n; i++) { for (int j = 0; j < 3; j++) { for (int k = 0; k < 3; k++) if (a[i-1]+j != a[i]+k) { f[i][k] = min(f[i][k], f[i-1][j] + k*b[i]); } } } ll ans = 2e18; for (int i = 0; i < 3; i++) { ans = min(ans, f[n][i]); } cout << ans << endl; } return 0; }