题目分析:
- 经典的流水作业调度问题
- 使用 Johnson算法,设 N1为a<b的集合,N2为a>=b的集合,将N1按照a非减序排序,N2按照b非增序排列,则N1作业接N2作业为最优决策
Code:
#include <bits/stdc++.h>
using namespace std;
#define maxn 1010
int n,a[maxn],b[maxn],ans[maxn];
struct node {
int w,id;
}e[maxn];
inline int read_() {
int x=0,f=1;
char c=getchar();
while(c<'0'||c>'9') {
if(c=='-') f=-1;
c=getchar();
}
while(c>='0'&&c<='9') {
x=(x<<1)+(x<<3)+c-'0';
c=getchar();
}
return x*f;
}
inline bool cmp_(node aa,node bb) {
return aa.w < bb.w;
}
inline void work_() {
for(int i=1;i<=n;++i) {
e[i].w=min(a[i],b[i]);
e[i].id=i;
}
sort(e+1,e+1+n,cmp_);
int l=0,r=n+1;
for(int i=1;i<=n;++i) {
if(e[i].w==a[e[i].id]) {
ans[++l]=e[i].id;
}
else {
ans[--r]=e[i].id;
}
}
l=0,r=0;
for(int i=1;i<=n;++i) {
int u=ans[i];
l+=a[u];
if(r<l) r=l;
r+=b[u];
}
printf("%d\n",r);
for(int i=1;i<=n-1;++i) {
printf("%d ",ans[i]);
}
printf("%d",ans[n]);
}
void readda_() {
n=read_();
for(int i=1;i<=n;++i) {
a[i]=read_();
}
for(int i=1;i<=n;++i) {
b[i]=read_();
}
work_();
}
int main() {
freopen("a.txt","r",stdin);
readda_();
return 0;
}