Shop scheduling problem

topic

Shop scheduling problem:

For each item of three ways:

1. A used time <B elapsed time. Called a class
2. A used time> B elapsed time. Referred to as second-class
3. And A, B equal to three cases. Called three

The problem into it, there is a class B is occupying time increases, the second category B is occupied with less time.

B is certainly the first change after more than fewer, to make time to spend a minimum.

Then consider the internal processing sequence in each case.

Equal without regard. No effect on the order.

First, the product is a waste of time to start the first time in A, we want to reduce it, so that a small to large.

Then when the last time wasted is time remaining in B, A, and more than that, so make large B in B in the process of first ran, ran the last little progress.

After consideration of the processing sequence can direct simulation.

#include <bits/stdc++.h>
#define N 100103
using namespace std;
int n, ans, delta;
struct product {
    int a, b, belong, id;
}data[N];       
bool cmp(product s, product b)
{               
    if (s.belong != b.belong)
        return s.belong < b.belong;
    if (s.belong == 3)
        return s.b > b.b;
    if (s.belong == 1) // 先开始的a一定要最小
        return s.a < b.a;
}               
int main()      
{               
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &data[i].a), data[i].id = i;
    for (int i = 1; i <= n; i++)
        scanf("%d", &data[i].b);
    for (int i = 1; i <= n; i++)//分类
    {   
        if (data[i].a == data[i].b);
            data[i].belong = 2;
        if (data[i].a < data[i].b)//先让b的时间长一些 
            data[i].belong = 1;
        if (data[i].a > data[i].b)
            data[i].belong = 3;
    }   
    sort(data + 1, data + 1 + n, cmp);
    for (int i = 1; i <= n; i++)//delta是指B中时间和A中时间差
    {   
        ans += data[i].a;
        delta -= data[i].a;
        delta = max(delta, 0);
        delta += data[i].b;
    }
    ans += delta;
    printf("%d\n", ans);
    for (int i = 1; i <= n; i++)
        printf("%d ", data[i].id);
    return 0;
}