USACO-Section2.1 Sorting a Three-Valued Sequence【数学公式】
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题目描述:
排序是一种很频繁的计算任务。现在考虑最多只有三值的排序问题。一个实际的例子是,当我们给某项竞赛的优胜者按金银铜牌排序的时候。在这个任务中可能的值只有三种1,2和3。我们用交换的方法把他排成升序的。
写一个程序计算出,给定的一个1,2,3组成的数字序列,排成升序所需的最少交换次数。(翻译来源:NOCOW)
INPUT FORMAT:
(file sort3.in)
第一行:
奖牌个数N (1 <= N <= 1000)
第 2行到第N+1行:
每行一个数字,表示奖牌。共N行。(1..3)
OUTPUT FORMAT:
(file sort3.out)
共一行,一个数字。表示排成升序所需的最少交换次数。
SAMPLE INPUT
9
2
2
1
3
3
3
2
3
1
SAMPLE OUTPUT
4
解题思路:
先遍历一遍数据,记录1,2,3的个数,再遍历一遍,记录c12(2在1的位置,后面同理)、c13、c21、c23、c32、c31、c32的个数,先计算只需要交换一次的,即min(c12,c21)+min(c13,c31)+min(c23+c32),再加上需要交换两次的,(abs(c13-c31)+abs(c23-c32)+abs(c12-c21))/3*2,就是剩下的没有归位的数字乘3除以2,即可得到结果。
#include<stdio.h>#include<string.h>#include<math.h>#include<stdlib.h>int n,a[1005];int min(int a,int b){ return a>b?b:a;}int main(){ FILE *fin = fopen ("sort3.in", "r"); FILE *fout = fopen ("sort3.out", "w"); int i,c1=0,c2=0,c3=0,c12=0,c21=0,c13=0,c31=0,c23=0,c32=0; fscanf(fin,"%d",&n); for(i=0;i<n;i++){ fscanf(fin,"%d",&a[i]); if(a[i]==1)c1++; if(a[i]==2)c2++; if(a[i]==3)c3++; } for(i=0;i<c1;i++){ if(a[i]==2)c12++; if(a[i]==3)c13++; } for(i=c1;i<c1+c2;i++){ if(a[i]==1)c21++; if(a[i]==3)c23++; } for(i=c1+c2;i<n;i++){ if(a[i]==1)c31++; if(a[i]==2)c32++; } int sum=min(c13,c31)+min(c12,c21)+min(c23,c32)+(abs(c13-c31)+abs(c23-c32)+abs(c12-c21))/3*2; fprintf(fout,"%d\n",sum); exit(0);}
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