解题思路:f[i] 表示包含i点的最大利润
注意事项:
参考代码:
#include<iostream> #include<algorithm> #include<cstring> #include<cmath> using namespace std; const int N = 1e2 + 10; int f[N],m[N],p[N],n,t,k,pre,ans; int main() { cin >> t; while(t--){ cin >> n >> k; memset(f,0,sizeof f); memset(m,0,sizeof m); memset(p,0,sizeof p); for(int i = 1; i <= n; i ++) cin >> m[i]; for(int i = 1; i <= n; i ++) { cin >> p[i]; f[i] = p[i]; } for(int i = 1; i <= n; i ++){ pre = i - 1; while(m[i] - m[pre] <= k && pre > 0) pre --; f[i] = max(f[i - 1],f[pre] + p[i]); } ans = 0; for(int i = 1; i <= n; i ++) ans = max(ans,f[i]); cout << ans <<endl; } return 0; }
另外一种解法(有两项测试数据未通过):
#include<iostream> #include<algorithm> #include<cstring> #include<cmath> using namespace std; const int N = 1e2 + 10; int f[N],r[N],m[N],p[N],n,t,k; int main() { freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); cin >> t; while(t--){ cin >> n >> k; memset(f,0,sizeof f); memset(m,0,sizeof m); memset(p,0,sizeof p); memset(r,0,sizeof r); for(int i = 1; i <= n; i ++) cin >> m[i]; for(int i = 1; i <= n; i ++) { cin >> p[i]; f[i] = p[i]; r[i] = p[i]; } for(int i = 1; i <= n; i ++) for(int j = n; j >= 1; j --){ if(abs(m[j] - m[i]) <= k) continue; f[j] = max(f[j], f[i] + p[j]); } for(int i = n; i >= 1; i --) for(int j = 1; j <= n; j ++){ if(abs(m[j] - m[i] ) <= k) continue; r[j] = max(r[j], r[i] + p[j]); } cout << ((f[n] > r[1])? f[n]:r[1] )<< endl; } return 0; }
0.0分
1 人评分