加工生产调度

根本不会。。。

题解：

https://www.cnblogs.com/f91og/p/6024706.html

https://blog.csdn.net/liufeng_king/article/details/8678316

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<vector>
 5 using namespace std;
 6 #define fi first
 7 #define se second
 8 #define mp make_pair
 9 #define pb push_back
10 typedef long long ll;
11 typedef unsigned long long ull;
12 typedef pair<int,int> pii;
13 struct Q
14 {
15     int x,y,n;
16 }q[1010],ans[1010];
17 bool operator<(const Q &a,const Q &b)
18 {
19     return min(a.x,a.y)<min(b.x,b.y);
20 }
21 int tl,tr,n;
22 int main()
23 {
24     int i,j,t;
25     scanf("%d",&n);tl=0;tr=n+1;
26     for(i=1;i<=n;i++)    scanf("%d",&q[i].x);
27     for(i=1;i<=n;i++)    scanf("%d",&q[i].y),q[i].n=i;
28     for(i=1;i<=n;++i)
29         for(j=i+1;j<=n;++j)
30             if(q[j]<q[i])
31                 swap(q[i],q[j]);
32     for(i=1;i<=n;i++)
33     {
34         t=min(q[i].x,q[i].y);
35         if(t==q[i].x)    ans[++tl]=q[i];
36         else    ans[--tr]=q[i];
37     }
38     int tm=0,an=0;
39     for(i=1;i<=n;i++)
40     {
41         tm+=ans[i].x;
42         an=max(an,tm)+ans[i].y;
43     }
44     printf("%d
",an);
45     return 0;
46 }

http://210.33.19.103/contest/966/problem/2

（玄学数据，不会字典序，然而数据也不是字典序）

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #include<vector>
 5 using namespace std;
 6 #define fi first
 7 #define se second
 8 #define mp make_pair
 9 #define pb push_back
10 typedef long long ll;
11 typedef unsigned long long ull;
12 typedef pair<int,int> pii;
13 struct Q
14 {
15     int x,y,n;
16 }q[1010],ans[1010];
17 bool operator<(const Q &a,const Q &b)
18 {
19     return min(a.x,a.y)<min(b.x,b.y);
20 }
21 int tl,tr,n;
22 int main()
23 {
24     int i,j,t;
25     scanf("%d",&n);tl=0;tr=n+1;
26     for(i=1;i<=n;i++)    scanf("%d",&q[i].x);
27     for(i=1;i<=n;i++)    scanf("%d",&q[i].y),q[i].n=i;
28     for(i=1;i<=n;++i)
29         for(j=i+1;j<=n;++j)
30             if(q[j]<q[i])
31                 swap(q[i],q[j]);
32     for(i=1;i<=n;i++)
33     {
34         t=min(q[i].x,q[i].y);
35         if(t==q[i].x)    ans[++tl]=q[i];
36         else    ans[--tr]=q[i];
37     }
38     int tm=0,an=0;
39     for(i=1;i<=n;i++)
40     {
41         tm+=ans[i].x;
42         an=max(an,tm)+ans[i].y;
43     }
44     printf("%d
",an);
45     for(i=1;i<=n;i++)    printf("%d ",ans[i].n);
46     return 0;
47 }