算法作业

#include<bits/stdc++.h>
using namespace std;
#define MAXN 911
#define Inf 0x3f3f3f3f
struct MatrixDimension
{
    int n,m;           ///表示矩阵是n*m维的
} Di[MAXN];

void TheMain();
void print(int **a, int **b, int n, int m, bool f);

void MatrixChainMain();
void MatrixChain(MatrixDimension MD[], int **s, int **bestnum, int n);
void print1(int **s, int l, int r);

void LCSlengthMain();
void LCSlength(int n, int m, char *s1, char *s2, int **c, int **b);
void print2(char *s1, char *s2, int nr, int mr, int **c);

void KnapsackMain();
void Knapsack(int **bestValue, int n, int *weight, int *value, int Capacity);
void print3(int **bv, int *w, int c, int n);
///-----------------------------------------------------------------
int main()
{
    TheMain();
    cout<<"程序已结束！"<<endl;
    return 0;
}
///-----------------------------------------------------------------
void TheMain()
{
    int flag;
    cout<<"--------------------------------------------------------------------------------------------"<<endl;
    cout<<"                                         主菜单                                             "<<endl;
    cout<<"功能介绍：演示矩阵连乘、最长公共子序列、0-1背包的动态规划。                                 "<<endl;
    cout<<"输入数字1进入矩阵连乘界面，输入2进入最长公共子序列界面，输入3进入0-1背包界面,输入0结束程序。"<<endl;
    cin>>flag;
    while(flag != 0 && flag != 1 && flag != 2 && flag != 3)
    {
        cout<<"输入错误，请按以下要求重新输入!"<<endl;
        cout<<"输入数字1进入矩阵连乘界面，输入2进入最长公共子序列界面，输入3进入0-1背包界面,输入0结束程序。"<<endl;
        cin>>flag;
    }
    cout<<"3秒后进入新界面..."<<endl;
    int t1 = time(NULL), t2 = time(NULL);
    while(t2- t1 < 3)
        t2 = time(NULL);
    system("cls");
    if(flag == 1)
        MatrixChainMain();
    else if(flag == 2)
        LCSlengthMain();
    else if(flag == 3)
        KnapsackMain();
    else
        return;
}
void print(int **a, int **b, int n, int m, bool f)///f = 1代表矩阵a,b相同，否则不同
{
    cout<<"等待3秒后输出"<<endl;
    int t1 = time(NULL);
    while(time(NULL) - t1 < 3) {}
    system("cls");
    if(f)
        for(int i = 0; i <= n; i++)
        {
            for(int j = 0; j <= m; j++)
                cout<<left<<fixed<<setw(5)<<a[i][j];
            cout<<endl;
        }
    else
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(i <= j)
                    cout<<left<<fixed<<setw(11)<<a[i][j];
                else
                    cout<<left<<fixed<<setw(11)<<b[j][i];

            }
            cout<<endl;
        }
}

///------------------------------------------------------矩阵连乘
void MatrixChainMain()
{
    cout<<"-------------------------------------------------"<<endl;
    cout<<"             矩阵连乘界面                        "<<endl;
    int n, i;
    cout<<"输入矩阵个数n = ";
    cin>>n;
    cout<<"输入"<<n<<"个矩阵的维数:";
    for(i = 0; i < n; i++)
        cin>>Di[i].n>>Di[i].m;
    int **s = new int*[n];
    for(int i = 0; i < n; i++)
        s[i] = new int[n];
    int **bm = new int*[n];
    for(int i = 0; i < n; i++)
        bm[i] = new int[n];
    MatrixChain(Di, s, bm, n);
    cout<<"矩阵相乘的次序为：";
    print1(s,0,n-1);
    cout<<endl;
    int flag;
    cout<<"输入0返回主菜单，输入1重新演示,输入2进入最长公共子序列界面，输入3进入0-1背包界面,输入4结束。"<<endl;
    cin>>flag;
    while(flag != 0 && flag != 1 && flag != 2 && flag != 3 &&flag != 4)
    {
        cout<<"输入错误，请按以下要求重新输入!"<<endl;
        cout<<"输入0返回主菜单，输入1重新演示,输入2进入最长公共子序列界面，输入3进入0-1背包界面,输入4结束。"<<endl;
        cin>>flag;
    }
    cout<<"3秒后进入新界面..."<<endl;
    int t1 = time(NULL), t2 = time(NULL);
    while(t2- t1 < 3)
        t2 = time(NULL);
    system("cls");
    if (flag == 0)
        TheMain();
    else if(flag == 1)
        MatrixChainMain();
    else if(flag == 2)
        LCSlengthMain();
    else if(flag == 3)
        KnapsackMain();
    else
        return;
}
void MatrixChain(MatrixDimension MD[], int **s, int **bestnum, int n)
{
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++)
        {
            if(i == j)
                bestnum[i][j] = 0;
            else
            {
                bestnum[i][j] = -1;
                s[i][j] = -1;
            }
        }
    print(bestnum, s, n, n, 0);
    for(int r = 1; r < n; r++)    ///r代表第r条对角线
    {
        for(int i = 0; i < n - r; i++)
        {
            int k = i + r;
            bestnum[i][k] = Inf;
            for(int j = i; j < k; j++) ///接下来求从第i个矩阵到第k个矩阵的最优值及最优解
            {
                int temp = bestnum[i][j] + bestnum[j+1][k] + MD[i].n * MD[j].m * MD[k].m;
                if(bestnum[i][k] >= temp)
                {
                    bestnum[i][k] = temp;
                    s[i][k] = j;
                }
            }
        }
        print(bestnum, s, n, n, 0);
    }
}
void print1(int **s, int l, int r)
{
    if(l >= r)
    {
        cout<<"A"<<l;
        return ;
    }
    int temp = s[l][r];
    cout<<"(";
    print1(s,l,temp);
    print1(s,temp+1,r);
    cout<<")";
}
///-------------------------------------------------最长公共子序列
void LCSlengthMain()
{
    cout<<"-------------------------------------------------"<<endl;
    cout<<"             最长公共子序列界面                  "<<endl;
    int n, m;
    cout<<"输入序列1的长度n = ";
    cin>>n;
    cout<<"输入序列2的长度m = ";
    cin>>m;
    char *s1 = new char[n+1];
    char *s2 = new char[m+1];
    cout<<"输入第一个序列的元素:";
    for(int i = 1; i <= n; i++)
        cin>>s1[i];
    cout<<"输入第二个序列的元素:";
    for(int i = 1; i <= m; i++)
        cin>>s2[i];
    int **c = new int*[n+1]; ///c[i][j]存储s1[i],s2[j]的最长公共子序列长度
    int **b = new int*[n+1]; ///b[i][j]表示c[i][j]是由哪个子问题得来的，作用是方便输出，有没有无所谓。
    for(int i = 0; i <= n; i++)
    {
        c[i] = new int[m+1];
        b[i] = new int[m+1];
    }
    LCSlength(n, m, s1, s2, c, b);
    cout<<"最长子序列长度为:"<<c[n][m]<<endl;
    cout<<"最长子序列为:";
    print2(s1, s2, n, m, c);
    cout<<endl;
    int flag;
    cout<<"输入0返回主菜单，输入1重新演示,输入2进入矩阵连乘界面，输入3进入0-1背包界面,输入4结束。"<<endl;
    cin>>flag;
    while(flag != 0 && flag != 1 && flag != 2 && flag != 3 &&flag != 4)
    {
        cout<<"输入错误，请按以下要求重新输入!"<<endl;
        cout<<"输入0返回主菜单，输入1重新演示,输入2进入矩阵连乘界面，输入3进入0-1背包界面,输入4结束。"<<endl;
        cin>>flag;
    }
    cout<<"3秒后进入新界面..."<<endl;
    int t1 = time(NULL), t2 = time(NULL);
    while(t2- t1 < 3)
        t2 = time(NULL);
    system("cls");
    if (flag == 0)
        TheMain();
    else if(flag == 2)
        MatrixChainMain();
    else if(flag == 1)
        LCSlengthMain();
    else if(flag == 3)
        KnapsackMain();
    else
        return;
}
void LCSlength(int n, int m, char *s1, char *s2, int **c, int **b)
{
    for(int i = 0; i <= n; i++)
        for(int j = 0; j <= m; j++)
        {
            if(i == 0 || j == 0)
                c[i][j] = 0;
            else
                c[i][j] = -1;
        }
    print(c, c, n, m, 1);
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {
            if(s1[i] == s2[j])
            {
                c[i][j] = c[i-1][j-1] + 1;
                b[i][j] = 0;
            }
            else
            {
                if(c[i-1][j] > c[i][j-1])
                {
                    c[i][j] = c[i-1][j];
                    b[i][j] = 1;
                }
                else
                {
                    c[i][j] = c[i][j-1];
                    b[i][j] = 2;
                }
            }
        }
        print(c, c, n, m, 1);
    }
}
void print2(char *s1, char *s2, int nr, int mr, int **c)
{
    if(nr == 0 || mr == 0)
        return;
    if(c[nr][mr] == c[nr-1][mr-1] + 1)
    {
        print2(s1,s2,nr-1,mr-1,c);
        cout<<s1[nr];
    }
    else
    {
        if(c[nr][mr] == c[nr-1][mr])
            print2(s1,s2,nr-1,mr,c);
        else
            print2(s1,s2,nr,mr-1,c);
    }
}


///--------------------------------------------------------0-1背包
void KnapsackMain()
{
    cout<<"-------------------------------------------------"<<endl;
    cout<<"                0-1背包界面                      "<<endl;
    int c;
    cout<<"输入背包容量:";
    cin>>c;
    int n;
    cout<<"输入物品数量:";
    cin>>n;
    int *w = new int[n];
    cout<<"输入每个物品的重量:";
    for(int i = 0; i < n; i++)
        cin>>w[i];
    int *v = new int[n];
    cout<<"输入每个物品的价值:";
    for(int i = 0; i < n; i++)
        cin>>v[i];
    int **bv = new int*[n];
    for(int i = 0; i < n; i++)
        bv[i] = new int[c+1];
    Knapsack(bv, n, w, v, c);
    cout<<"背包可装入的最大价值为:"<<bv[0][c]<<endl;
    cout<<"背包装入物品的一种方案为:"<<endl;
    print3(bv,w,c,n);

    int flag;
    cout<<"输入0返回主菜单，输入1重新演示,输入2进入最长公共子序列界面，输入3进入矩阵连乘界面,输入4结束。"<<endl;
    cin>>flag;
    while(flag != 0 && flag != 1 && flag != 2 && flag != 3 &&flag != 4)
    {
        cout<<"输入错误，请按以下要求重新输入!"<<endl;
        cout<<"输入0返回主菜单，输入1重新演示,输入2进入最长公共子序列界面，输入3进入矩阵连乘界面,输入4结束。"<<endl;
        cin>>flag;
    }
    cout<<"3秒后进入新界面..."<<endl;
    int t1 = time(NULL), t2 = time(NULL);
    while(t2- t1 < 3)
        t2 = time(NULL);
    system("cls");
    if (flag == 0)
        TheMain();
    else if(flag == 3)
        MatrixChainMain();
    else if(flag == 2)
        LCSlengthMain();
    else if(flag == 1)
        KnapsackMain();
    else
        return;
}
void Knapsack(int **bestValue, int n, int *weight, int *value, int Capacity)
{
    int jMax = (weight[n-1] - 1) < Capacity ? (weight[n-1] - 1) : Capacity;
    for(int i = 0; i < n; i++)
        for(int j = 0; j <= Capacity; j++)
        {
            if(i == n - 1 && j <= jMax)
                bestValue[i][j] = 0;
            else if(i == n - 1 && j > jMax)
                bestValue[i][j] = value[n - 1];
            else
                bestValue[i][j] = -1;
        }
    print(bestValue, bestValue, n - 1, Capacity, 1);
    for(int i = n - 2; i > 0; i--)
    {
        for(int j = 0; j <= Capacity; j++)
        {
            int temp1 = j - weight[i];
            if(temp1 < 0)
                bestValue[i][j] = bestValue[i+1][j];
            else
            {
                int temp2 = bestValue[i+1][temp1] + value[i];
                bestValue[i][j] = temp2 > bestValue[i+1][j] ? temp2 : bestValue[i+1][j];
            }
        }
        print(bestValue, bestValue, n - 1, Capacity, 1);
    }
    bestValue[0][Capacity] = bestValue[1][Capacity];
    int temp = Capacity - weight[0];
    if(temp >= 0)
        bestValue[0][Capacity] = bestValue[0][Capacity] > ( bestValue[1][temp] + value[0]) ? bestValue[0][Capacity] : ( bestValue[1][temp] + value[0]);
    print(bestValue, bestValue, n - 1, Capacity, 1);
}
void print3(int **bv, int *w, int c, int n)
{
    int j = c;
    for(int i = 0; i < n - 1; i++)
    {
        if(bv[i][j] == bv[i+1][j])
            continue;
        cout<<"物品"<<i + 1<<" ";
        j = j - w[i];
    }
    if(bv[n - 1][j] == 0)
        cout<<endl;
    else
        cout<<"物品"<<n<<endl;
}

来源：CSDN

作者：Only*

链接：https://blog.csdn.net/weixin_43040863/article/details/103715372