Bruteforce

Problem Description

Given three arraies A[],B[],C[], each contains N non-negative integers.You are asked to maxmize the vale:V=max(A[i]+B[j]+C[k]), where 0<i,j,k<=N and i!=j and j!=k and i!=k.

Input

Each case contains 4 lines,
the first line contains an integer N( 3<=N<=10000 ) ,
the second line contains N integers representing array A[],
the third line contains N integers representing array B[],
the fourth line contains N integers representing array C[].

Output

Each case contains a number seperately: the answer V.

Sample Input

31 2 33 2 13 2 1

Sample Output

8

由于前一阵子在帮着做电子综合设计的破东西，竟然把我挚爱的ACM给遗忘了一阵子，真是该死啊！！回到杭电开始做题，今天本打算K掉DIY里面的一个比赛的，居然花了一个小时的时候做了一道水题，真是不应该啊，不过也巩固了DFS的用法，嘿嘿，几天不做手变得好生疏啊。。。。

# include<stdio.h># include<string.h>int a[10001],b[10002],c[10003];intid1[10001],id2[10002],id3[10003];int max(int i,int j,int k){i=i>j?i:j;k=k>i?k:i;return k;}int max2(int i,int j){return i>j?i:j;}int cmp1(const void *d,const void *e){return a[*((int*)e)]-a[*((int*)d)];}int cmp2(const void *d,const void *e){return b[*((int*)e)]-b[*((int*)d)];}int cmp3(const void *d,const void *e){return c[*((int*)e)]-c[*((int*)d)];}void print(int *a,int n){int i;for(i=0;i<n;i++)printf("%d ",a[i]);printf("\n");}int work(int i,int j,int k){if(id1[i]!=id2[j]&&id1[i]!=id3[k]&&id2[j]!=id3[k])return a[id1[i]]+b[id2[j]]+c[id3[k]];if(id1[i]==id2[j])return max2(work(i+1,j,k),work(i,j+1,k));if(id1[i]==id3[k])return max2(work(i+1,j,k),work(i,j,k+1));if(id3[k]==id2[j])return max2(work(i,j+1,k),work(i,j,k+1));}void main(){int n,i,j;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++)scanf("%d",&a[i]);for(i=0;i<n;i++)scanf("%d",&b[i]);for(i=0;i<n;i++){scanf("%d",&c[i]);id1[i]=i,id2[i]=i,id3[i]=i;}qsort(id1,n,sizeof(id1[0]),cmp1);qsort(id2,n,sizeof(id2[0]),cmp2);qsort(id3,n,sizeof(id3[0]),cmp3);printf("%d\n",work(0,0,0));}}