算法导论 16-1.1活动选择问题 动态规划解

#include <iostream>#include <string>using namespace std;#define N 11void dynamic_activity_selector(int c[N+1][N+1], int rec[N+1][N+1], int s[N+1], int f[N+1]){    for(int i = 1; i <= N; i++)        for(int j = 2; j <= N; j++)        {            if(i >= j)                c[i][j] = 0;            else            {                for(int k = i+1; k <= j-1; k++)                    if(f[i] <= s[k] && f[k] <= s[j])//保证k的兼容性                    {                        if(c[i][j] < c[i][k] + c[k][j] + 1)//递归计算公式                        {                            c[i][j] = c[i][k] + c[k][j] + 1;                            rec[i][j] = k;                        }                    }            }        }}void print(int rec[N+1][N+1], int i, int j)//自顶向下(从后往前)递归寻找最优k值{    if(i<=j)    {        print(rec, i, rec[i][j]);        if(rec[i][j] != 0)//k值非零时,后序输出            cout<<"a"<<rec[i][j]<<" ";    }}int main(){    int s[N+1] = {-1,1,3,0,5,3,5,6,8,8,2,12};//起始时间    int f[N+1] = {-1,4,5,6,7,8,9,10,11,12,13,14};//结束时间    int c[N+1][N+1] = {0};//最大兼容子集中的活动数    int rec[N+1][N+1] = {0};//用于标记最优k值    dynamic_activity_selector(c,rec,s,f);    for(int i = 1; i < 12; i++)    {        for(int j = 1; j < 12; j++)            cout << c[i][j]<<" ";        cout<<endl;    }    cout<<"最大子集数为: "<<c[1][N] + 2<<endl;//a1和a未算在内,要加上    cout<<"最大子集为: {a1 ";    print(rec, 1, N);    cout<<"a"<<N<<"}";}

输出:
最大子集数为: 4
最大子集为: {a1 a4 a8 a11}