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poj 3270 Cow Sorting

http://poj.org/problem?id=3270

这道题就是给你一个无序序列转换成有序序列需要花费的代价最小,交换a和b代价为a+b;

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #define maxn 20000
 5 using namespace std;
 6 
 7 int min1=100000;
 8 int n;
 9 bool vis[maxn];
10 struct node
11 {
12   int num;
13   int id;
14   bool operator <(const node &a)const
15   {
16       return num<a.num;
17   }
18 }p[maxn*2];
19 
20 int deal()
21 {
22     int ans=0;
23     for(int i=0; i<n; i++)
24     {
25         if(!vis[i])
26         {
27             vis[i]=true;
28             int sum=p[i].num;
29             int m1=p[i].num;
30             int t1=1;
31             int id=p[i].id;
32             while(id!=i)
33             {
34                 vis[id]=true;
35                 sum+=p[id].num;
36                 m1=min(m1,p[id].num);
37                 id=p[id].id;
38                 t1++;
39             }
40             ans+=min(sum+(t1-2)*m1,sum+m1+(t1+1)*min1);
41         }
42     }
43     return ans;
44 }
45 
46 int main()
47 {
48     while(scanf("%d",&n)!=EOF)
49     {
50         memset(vis,false,sizeof(vis));
51         for(int i=0; i<n; i++)
52         {
53             scanf("%d",&p[i].num);
54             p[i].id=i;
55             min1=min(min1,p[i].num);
56         }
57         sort(p,p+n);
58         printf("%d\n",deal());
59     }
60     return 0;
61 }
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