贪心法

1.单源最短路径问题

 

#include<stdio.h>

#include<iostream>

#include<cmath>

using  namespace std;

#define M   1000

int Cost[20][20];

int n;

int Distance[20];

bool s[20];

int pr[20];

int source;

void Dijkstra()

{

    int i, j;

    //初始化

    for(i=1; i<=n; i++)

    {

        Distance[i] = Cost[source][i];

        s[i] = false;

        if(Distance[i]==M)

        {

           pr[i]=0;

        }

        else

        {

            pr[i]=source;

        }

    }

    Distance[source] = 0;

    s[source] = true;

    for(i=1; i < n; i++)

    {//每循环一次,求得一个最短路径

       int temp=M;

       int u=source;

        for (j=1; j <= n; j++) //求离出发点最近的顶点

           if((!s[j]) && Distance[j]<temp)

           {

               temp = Distance [j];

               u = j;

           }

        s[u] =true;

        for(j=1; j<=n; j++) //修改递增路径序列(集合)

        {

        if((!s[j])&& Cost[u][j]<M)

        { //对还未求得最短路径的顶点

            //求出由最近的顶点 直达各顶点的距离

            int newdis = Distance[u] +Cost[u][j];

            // 如果新的路径更短,就替换掉原路径

            if(Distance[j] > newdis)

            {

                Distance[j] = newdis;

                pr[j] = u;

            }

        }

        }

    }

}

void Traceback(int i)

{

    if(source == i)

    {

        printf("%d",i);

        return;

    }

    Traceback(pr[i]);

    cout<<"->"<<i;

}

int main()

{

    cout<<"请输入个数:"<<endl;

    cin>>n;

    int i,j;

    source = 1;//源点为1

    cout<<"请输入代价矩阵:"<<endl;

    for(i=1; i<=n; i++)

    {

        for(j=1; j<=n; j++)

        {

           scanf("%d",&Cost[i][j]);

        }

    }

    Dijkstra();

 

    for(i=2; i<=n; i++)

    {

        printf("从1到点%d的最短路径长度:%d,路径:",i,Distance[i]);

        Traceback(i);

        printf("\n");

    }

 system("pause");

    return 0;

}

 

2.多机调度问题

假定有7个独立作业,所需处理时间分别为{2,14,4,16,6,5,3},由三台机器M1,M2,M3加工。按照贪心算法所需总加工时间为17.

#include<stdio.h>

#include<algorithm>

#include<iostream>

using namespace std;

struct work

{

    int time;

    int num;

}homework[20];

void multimachine(work homework[20],int n,int m)

{

    int h[20][20]={0};

    int d[20];

    int real[20];

    int sum,i,j,k;

    for(i=1;i<=m;i++)

    {

        h[i][1]=homework[i].num;

        real[i]=1;

        d[i]=homework[i].time;

    }

    for(i=m+1;i<=n;i++)

    {

        for(j=1,k=2;k<=m;k++)

           if(d[k]<d[j])

               j=k;

           real[j]++;

           h[j][real[j]]=homework[i].num;

           d[j]=d[j]+homework[i].time;

    }

    for(i=1;i<=m;i++)

    {

        printf("机器%d处理:",i);

        for(j=1;h[i][j]>0;j++)

        {

           printf("作业%d ",h[i][j]);

        }

        printf("\n");

    }

    for(i=1;i<=m;i++)

    {

        sum=d[i]>d[i+1]?d[i]:d[i+1];

    }

        printf("完成时间为:%d\n",sum);

}

bool cmp(work a,work b)

{

    return a.time>b.time;

}

int main()

{

    int m;

    printf("机器个数:");

    cin>>m;

    int n;

    printf("作业个数:");

    cin>>n;

    printf("各作业完成所需时间:\n");

    int i;

    for(i=1;i<=n;i++)

    {

        scanf("%d",&homework[i].time);

        homework[i].num=i;

    }

    sort(homework+1,homework+10,cmp);

    multimachine(homework,n,m);

    system("pause");

    return 0;

}

posted @ 2019-06-17 08:50  夜游星  阅读(246)  评论(0编辑  收藏  举报