usaco2.12Sorting a Three-Valued Sequence

Sorting a Three-Valued Sequence
IOI'96 - Day 2

Sorting is one of the most frequently performed computational tasks. Consider the special sorting problem in which the records to be sorted have at most three different key values. This happens for instance when we sort medalists of a competition according to medal value, that is, gold medalists come first, followed by silver, and bronze medalists come last.

In this task the possible key values are the integers 1, 2 and 3. The required sorting order is non-decreasing. However, sorting has to be accomplished by a sequence of exchange operations. An exchange operation, defined by two position numbers p and q, exchanges the elements in positions p and q.

You are given a sequence of key values. Write a program that computes the minimal number of exchange operations that are necessary to make the sequence sorted.

PROGRAM NAME: sort3

INPUT FORMAT

Line 1:	N (1 <= N <= 1000), the number of records to be sorted
Lines 2-N+1:	A single integer from the set {1, 2, 3}

SAMPLE INPUT (file sort3.in)

9
2
2
1
3
3
3
2
3
1

OUTPUT FORMAT

A single line containing the number of exchanges required

SAMPLE OUTPUT (file sort3.out)

4

剛開始以爲是選擇排序 WA了

先把3放在它應該在的位置 並且交換的時候保證1在2的前面 同樣再把2換到它該在的位置

/*
    ID: your_id_here
    PROG: sort3
    LANG: C++
*/
#include <iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int nu[1010],y[1010];
int main()
{
    freopen("sort3.in","r",stdin);
    freopen("sort3.out","w",stdout);
    int i,j,k,n,s=0,x[5]={0},g=1,t;
    cin>>n;
    for(i = 1; i <= n ; i++)
    {
       cin>>nu[i];
       x[nu[i]]++;
    }
    x[2]+=x[1];
    x[3]+=x[2];
    for(i = 1; i <= x[2] ; i++)
    {
        if(nu[i]==3)
        {
            s++;
            k = i;
            for(j = x[2]+1 ; j <= x[3] ; j++)
                if(nu[j]==1)
                {
                    k = j;
                    break;
                }
            if(k==i)
            {
                for(j = x[2]+1 ; j <= x[3] ; j++)
                    if(nu[j]==2)
                    {
                        k = j;
                        break;
                    }
            }
            t = nu[i];
            nu[i] = nu[k];
            nu[k] = t;
        }
    }
    for(i = 1; i <= x[1] ; i++)
    {
        if(nu[i]==2)
        s++;
    }
    cout<<s<<endl;
    fclose(stdin);
    fclose(stdout);
    return 0;
}