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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