poj 3270

置换群~

所求即为每个循环群需用的最小值之和,即min(sum - curmin+curmin*(len-1),sum + curmin + Min*(len+1));

//============================================================================
// Name        : 3270.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

int a[10010], goal[10010], flag[10010], num[100010];
int curans, ans, Min, curmin, sum, len, t,com, n;


int main() {
	freopen("a.txt", "r", stdin);
	while(scanf("%d", &n)!=EOF){
		memset(a, 0, sizeof(a));
		memset(flag, 0, sizeof(flag));
		memset(goal, 0, sizeof(goal));
		memset(num, 0, sizeof(num));
		for(int i = 0;i < n;i++){
			scanf("%d", &a[i]);
			goal[i] = a[i];
		}
		sort(goal, goal+n);
		for(int i = 0;i < n;i++){
			num[goal[i]] = i;
		}
		Min = goal[0];
		ans = 0;
		for(int i = 0;i < n;i++){
			if(flag[i] == 1) continue;
			flag[i] = 1;
			if(goal[i] == a[i]){
				continue;
			}
			com = goal[i];
			sum = goal[i];
			t = a[i];
			curmin = goal[i];
			len = 1;
			while(t != com){
				flag[num[t]] = 1;
				sum += t;
				len++;
				if(t < curmin){
					curmin = t;
				}
				t = a[num[t]];
			}
			curans = sum - curmin+curmin*(len-1);
			if(curans > sum + curmin + Min*(len+1)){
				curans = sum + curmin + Min*(len+1);
			}
			ans += curans;
		}
		printf("%d\n", ans);
	}
	return 0;
}

  

posted @ 2011-07-23 20:06  KOKO's  阅读(323)  评论(0编辑  收藏  举报