該類問題可以使用貪心策略可以設(shè)計(jì)出較好的近似算法（并不是最優(yōu)解）

#include<stdio.h> #include<stdlib.h> #include<iostream> #include<algorithm> using namespace std;typedef struct Node {int hour;//記錄該作業(yè)的所需要完成的時(shí)間int number;//記錄該作業(yè)在原來的序列的順序 }node; bool cmp(node x,node y) {if(x.hour>y.hour)return true;elsereturn false; } void MultiMachine(Node node[], int machine_number, int *machine_operation, int *machine_time[],int job_number){int rear[job_number];//安排前機(jī)器數(shù)（機(jī)器的個(gè)數(shù)）個(gè)作業(yè)for (int i = 0; i < machine_number;i++){machine_time[i][0] = node[i].number;//將該作業(yè)放到二維數(shù)組中去rear[i] = 0;machine_operation[i] = node[i].hour;}cout<<"dsfdsjfklsdjflksdj";//安排剩余的job_number-machine_number個(gè)作業(yè)int i;int j;int k;for( i= machine_number;i<job_number;i++){for ( j = 0,k=1; k < machine_number;k++)//找到目前機(jī)器序列中使用時(shí)間最短的那臺(tái)機(jī)器{if(machine_operation[j]>machine_operation[k]){j = k;}}rear[j]++;machine_time[j][rear[j]] = node[i].number;machine_operation[j] += node[i].hour;}for (int i = 0; i < machine_number;i++){cout << "機(jī)器" << i + 1 << "處理：";for (int j = 0; machine_time[i][j]>0;j++){cout << "作業(yè)" << machine_time[i][j] << " ";}cout << "機(jī)器" << i << "的處理時(shí)間" << machine_operation[i] << endl;}} int main() {cout << "輸入待處理的作業(yè)個(gè)數(shù)：";int job_number;cin >> job_number;Node node[job_number];cout << "輸入作業(yè)需要處理的時(shí)間：";for (int i = 0; i < job_number;i++){cin >> node[i].hour;node[i].number = i + 1;}cout << "輸入機(jī)器的個(gè)數(shù)：";int machine_number;cin >> machine_number;sort(node, node + job_number, cmp);int machine_time[machine_number][job_number];//記錄每個(gè)機(jī)器上作業(yè)的序號(hào) int machine_operation[machine_number];//記錄每個(gè)機(jī)器上已經(jīng)的運(yùn)行時(shí)間for (int i = 0; i < machine_number;i++){machine_operation[i] = 0;}MultiMachine(node, machine_number, machine_operation, (int**)machine_time, job_number);}

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