[csp-201809-4]再卖菜 差分约束or记忆化搜索

Posted 2021-01-25 konjakjuruo

 

先更新第一个做法：差分约束

转化成最长路，求出的每一个解是满足差分方程的最小值

 

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int N=3100;
 5 struct node{
 6     int x,y,d,next;
 7 }a[2*N];
 8 int n,al,first[N],b[N],s[N];
 9 bool vis[N];
10 queue<int> q;
11 
12 void ins(int x,int y,int d)
13 {
14     al++;
15     a[al].x=x;a[al].y=y;a[al].d=d;
16     a[al].next=first[x];first[x]=al;
17 }
18 
19 void build_edge()
20 {
21     al=0;
22     memset(first,0,sizeof(first));
23     ins(0,2,2*b[1]);
24     ins(2,0,-(2*b[1]+1));
25     for(int i=2;i<=n-1;i++)
26     {
27         ins(i-2,i+1,3*b[i]);
28         ins(i+1,i-2,-(3*b[i]+2));
29     }
30     ins(n-2,n,2*b[n]);
31     ins(n,n-2,-(2*b[n]+1));
32     for(int i=1;i<=n;i++) ins(i-1,i,1);
33 }
34 
35 void spfa()
36 {
37     while(!q.empty()) q.pop();
38     memset(s,0,sizeof(s));
39     memset(vis,0,sizeof(vis));
40     s[0]=0;vis[0]=1;q.push(0);
41     while(!q.empty())
42     {
43         int x=q.front();q.pop();
44         for(int i=first[x];i;i=a[i].next)
45         {
46             int y=a[i].y;
47             if(s[y]<s[x]+a[i].d)
48             {
49                 s[y]=s[x]+a[i].d;
50                 if(!vis[y])
51                 {
52                     vis[y]=1;
53                     q.push(y);
54                 }
55             }
56         }
57         vis[x]=0;
58     }
59 }
60 
61 int main()
62 {
63     //freopen("a.in","r",stdin);
64     scanf("%d",&n);
65     for(int i=1;i<=n;i++) scanf("%d",&b[i]);
66     build_edge();
67     spfa();
68     for(int i=1;i<=n;i++)
69         printf("%d ",s[i]-s[i-1]);printf("
");
70     return 0;
71 }