1 #include<iostream> 2 using namespace std; 3 4 int n; //主题的数量 5 int L, C; //每节课的时间，常量C 6 int *time; //每个主题需要的时间 7 int *minLec; //主题1~i（1<=i<n)的最少讲课次数 8 int *minDis; //与主题相应的最小不满意指标 9 10 //计算不满意指标 11 int DI(int t) 12 { 13 if (t == 0) return 0; 14 else if (t <= 10)return -C; 15 else return (t - 10)*(t - 10); 16 } 17 18 19 //动态规划的算法实现 20 void DP() 21 { 22 int i, j; 23 int cost; 24 int sum; 25 minDis[0] = 0; 26 minLec[0] = 0; 27 for (i = 1; i <= n; i++) 28 { 29 minLec[i] = 30000; 30 sum = 0; 31 for (j = i-1; j>=0; j--) 32 { 33 sum += time[j + 1]; 34 if (sum > L) break; 35 if (minLec[j] + 1 > minLec[i]) continue;//如果j的次数+1大于i继续访问前面 36 cost = minDis[j] + DI(L - sum); 37 if (minLec[j] + 1 == minLec[i] && cost >= minDis[i]) continue;如果j的次数+1等于i，并且cost大于i的满意值，继续访问前面的 38 minDis[i] = cost; 39 minLec[i] = minLec[j] + 1; 40 } 41 } 42 } 43 44 45 void main() 46 { 47 cin >> n; 48 cin >> L >> C; 49 minDis = new int[n + 1]; 50 minLec = new int[n + 1]; 51 time = new int[n + 1]; 52 for (int i = 1; i <= n; i++) 53 cin >> time[i]; 54 DP(); 55 cout << "Minimum number of lectures:" << minLec[n]<<endl; 56 cout << "Total dissatisfaction index:" << minDis[n] << endl; 57 }
