算法实验3《动态规划算法实验》

1. 编写一个简单的程序，解决0-1背包问题。设N=5，C=10，w={2，2，6，5，4},v={6,3,5,4,6} 

#include<iostream>using namespace std;void knapsack(int *v, int *w, int c, int n, int**m){int jmax = (w[n] > c) ? c : w[n]-1;for (int j = 0; j <= jmax; j++)m[n][j] = 0;for (int j = w[n]; j <= c; j++)m[n][j] = v[n];for (int i = n - 1; i > 1; i--){jmax = (w[i] > c) ? c : w[i] - 1;for (int j = 0; j <= jmax; j++)m[i][j] = m[i + 1][j];for (int j = w[n]; j <= c; j++)m[i][j] = (m[i + 1][j] > m[i + 1][j - w[i]] + v[i])   ?   m[i + 1][j]  :  m[i + 1][j - w[i]] + v[i];}m[1][c] = m[2][c];if (c >= w[1])m[1][c] = (m[1][c] > m[2][c - w[1]] + v[1]) ? m[1][c] : m[2][c - w[1]] + v[1];}int main(){int n = 5;int c = 10;int w[6] ={ 0, 2, 2, 6, 5, 4 };int v[6] = { 0, 6, 3, 5, 4, 6 };int **m = new int*[6];for (int i = 0; i < 6; i++)m[i] = new int[6];knapsack(v, w, c, n, m);cout << " 最优值为  " << m[1][c] <<"\n最优解为取第  ";for (int i = 1; i < n; i++){if (m[i][c] != m[i + 1][c]){cout << i << "   ";c -= w[i];}}if (m[n][c])cout << n;cout << " 个物品";system("pause>nul");return 0;}

2. 合唱队形安排问题

【问题描述】N位同学站成一排，音乐老师要请其中的(N-K)位同学出列，使得剩下的K位同学排成合唱队形。合唱队形是指这样的一种队形：设K位同学从左到右依次编号为1，2…，K，他们的身高分别为T1，T2，…，TK，  则他们的身高满足T1<...<Ti>Ti+1>…>TK(1<=i<=K)。已知所有N位同学的身高，计算最少需要几位同学出列，可以使得剩下的同学排成合唱队形。

#include<iostream>using namespace std;//计算以list[i]为结尾的递增数组的最大长度length[i]void getLength(int n, double *list, int *length){length[0] = 0;for (int i = 1; i <= n; i++){length[i] = 0;for(int j=0;j<i;j++)if(list[i]>list[j] && length[i]<length[j]+1)length[i] = length[j] + 1;}}void change(double *list, int n)//把数组反过来{for (int i = 1, j = n; i < j; i++, j--){double s = list[i] + list[j];list[i] = s - list[i];list[j] = s - list[j];}}int main(){int n;cout << "输入n和n个正数\n";cin >> n;double *list = new double[n + 1];list[0] = 0;for (int i = 1; i <= n; i++)cin >> list[i];int *length1 = new int[n + 1], *length2 = new int[n + 1];getLength(n, list, length1);change(list, n);getLength(n, list, length2);int lengthmax = 0;//最大长度int index = 0;//最大长度对应的最大数的下标for (int i = 1; i <= n; i++){if (length1[i] + length2[n + 1 - i] > lengthmax){lengthmax = length1[i] + length2[n + 1 - i];index = i;}}cout << "合唱队形最大长度为" << lengthmax-1 << "\n最高者是第" <<index<<"个，高"<< list[index+1];system("pause>nul");return 0;}