Cow Sorting(DP之—置换群)
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Cow Sorting
Time Limit: 2000MS
Memory Limit: 65536K
Total Submissions: 5062
Accepted: 1861
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
Line 1: A single integer: N.
Lines 2..N+1: Each line contains a single integer: line i+1 describes the grumpiness of cow i.
Output
Line 1: A single line with the minimal time required to reorder the cows in increasing order of grumpiness.
Sample Input
3
2
3
1
Sample Output
7
Hint
2 3 1 : Initial order.
2 1 3 : After interchanging cows with grumpiness 3 and 1 (time=1+3=4).
1 2 3 : After interchanging cows with grumpiness 1 and 2 (time=2+1=3).
Source
参考博客
http://blog.csdn.net/woshi250hua/article/details/7648099
置换群的应用:
sum - min + (len - 1) * min和sum + min + (len + 1) * smallest之中较小的那个值。
#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;const int maxn=10000+10;int a[maxn],c[100010];int vis[maxn],n;const int inf=1<<30;int MIN,MAX,sum;int min(int a,int b){ return a<b? a:b;}int work() { int k,t,i,j,ans=0; for(i=1;i<=n;i++) { if(!vis[i]) { ans=0; j=i; t=a[i];k=0; while(!vis[j]) {k++; if(t>a[j]) t=a[j]; vis[j]=1; j=c[a[j]]; } //这里的K经常出现K=1的情况 ans+=min((k-2)*t,t+(k+1)*MIN); sum+=ans; //printf("k=%d,ans=%d\n",k,ans); }} return sum; }int main(){ int i,j; int ans; while(scanf("%d",&n)!=EOF) { sum=0; MIN=inf,MAX=0; memset(vis,0,sizeof(vis)); memset(c,0,sizeof(c)); for(i=1;i<=n;i++) {scanf("%d",&a[i]); c[a[i]]++; if(a[i]<MIN) MIN=a[i]; if(a[i]>MAX) MAX=a[i]; sum+=a[i];//轮换的最初情况是每个值都要被换一次 } //printf("min=%d max=%d",MIN,MAX); for(j=1;j<=MAX;j++) c[j]=c[j]+c[j-1];//求应有的位置 ans=work(); printf("%d\n",ans); } //system("pause"); return 0;}
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