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| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 6287 | Accepted: 2429 |
Description
Farmer John‘s N (1 ≤ N ≤ 10,000) cows are lined up to be milked in the evening. Each cow has a unique "grumpiness" level in the range 1...100,000. Since grumpy cows are more likely to damage FJ‘s milking equipment, FJ would like to reorder the cows in line so they are lined up in increasing order of grumpiness. During this process, the places of any two cows (not necessarily adjacent) can be interchanged. Since grumpy cows are harder to move, it takes FJ a total of X+Y units of time to exchange two cows whose grumpiness levels are X and Y.
Please help FJ calculate the minimal time required to reorder the cows.
Input
Output
Sample Input
3 2 3 1
Sample Output
7
Hint
Source
#include <iostream> #include <algorithm> #include <cstdio> using namespace std; #define INF 0x3f3f3f3f #define N 10010 int n; int mi; int b[N]; int vis[N]; struct Node { int ori,now; }a[N]; int main() { while(scanf("%d",&n)!=EOF) { mi=INF; for(int i=1;i<=n;i++) { scanf("%d",&a[i].ori); mi=min(a[i].ori,mi); b[i]=a[i].ori; } sort(b+1,b+n+1); for(int i=1;i<=n;i++) { a[i].now=lower_bound(b+1,b+n+1,a[i].ori)-b; } int ans=0; int len,sum,t1,t2; for(int i=1;i<=n;i++) { if(a[i].now!=i) { len=1; sum=0; while(a[i].now!=i) { len++; sum+=a[i].ori; swap(a[i],a[a[i].now]); } t1=sum+(len-1)*a[i].ori; t2=sum+(len-1)*mi+(a[i].ori+mi)*2; ans+=min(t1,t2); } } printf("%d\n",ans); } return 0; }
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原文地址：http://www.cnblogs.com/hate13/p/4430655.html
