回溯法:N后问题

#include

#include <cstdio>

using namespace std;

#define MAX(a, b) ((a) > (b) ? (a) : (b))

#define MAXSIZE 10

#define INF 999999999

struct Job_Info

{

int t1, t2;

}job[MAXSIZE];

int n;
//作业数量

int solu[MAXSIZE];
//调度次序

int optimalValue;
//最优值

int optimalSolu[MAXSIZE];
//最优解

void Swap(int &a, int &b)

{

int temp = a;

a = b;

b = temp;

}

//time1,time2分别为机器1和机器2的完成处理时间

//totalTime为已完成作业的时间和

void Backtrack(int k, int time1, int time2, int totalTime)

{

int i;

if (k >= n)

{

for (i = 0; i < n; i++)

{

optimalSolu[i] = solu[i];

}

optimalValue = totalTime;

}

else

{

for (i = k; i < n; i++)

{

//计算完成第i个任务后,两机器的时间

int nextTime1 = time1 + job[solu[i]].t1;

int nextTime2 = MAX(nextTime1, time2) + job[solu[i]].t2;

//已完成作业的时间和是否小于最优值

if (totalTime + nextTime2 < optimalValue)

{

Swap(solu[k], solu[i]);

Backtrack(k+1, nextTime1, nextTime2, totalTime+nextTime2);

Swap(solu[k], solu[i]);

}

}

}

}

int main(void)

{

int i;

for (i = 0; i < MAXSIZE; i++)

{

solu[i] = i;

}

while (scanf(“%d”, &n) != EOF)

{

for (i = 0; i < n; i++)

{

scanf(“%d%d”, &job[i].t1, &job[i].t2);

}

optimalValue = INF;

Backtrack(0, 0, 0, 0);

printf(“optimal value: %d\n”, optimalValue);

printf(“optimal solution: “);

for (i = 0; i < n-1; i++)

{

printf(“%d “, optimalSolu[i]+1);

}

printf(“%d\n”, optimalSolu[i]+1);

}

return 0;

}

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