【递归】【3月周赛1】【Problem B】

Problem B

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 59   Accepted Submission(s) : 27

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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

3
1 2 3
3 2 1
3 2 1

Sample Output

8

思路：

int dfs(int i,int j,int k)

{

if(A[i].pos!=B[j].pos&&A[i].pos!=C[k].pos&&B[j].pos!=C[k].pos)

return A[i].l+B[j].l+C[k].l;

if(A[i].pos==B[j].pos)  

        return max(dfs(i+1,j,k),dfs(i,j+1,k));  

        if(A[i].pos==C[k].pos)  

        return max(dfs(i+1,j,k),dfs(i,j,k+1));  

        if(C[k].pos==B[j].pos)  

        return max(dfs(i,j+1,k),dfs(i,j,k+1));  

}

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
struct node
{
	int l;int pos;
};
node A[10001],B[10001],C[10001];
int N;
int cmp(const void *i,const void *j)
{
	 node *ii=(node *)i,*jj=(node *)j;
	 return jj->l-ii->l;
}
void input()
{
	for(int i=1;i<=N;i++)
	{
		scanf("%d",&A[i].l);
		A[i].pos=i;
	}
	for(int i=1;i<=N;i++)
	{
		scanf("%d",&B[i].l);
		B[i].pos=i;
	}
	for(int i=1;i<=N;i++)
	{
		scanf("%d",&C[i].l);
		C[i].pos=i;
	}
	         qsort(A+1,N,sizeof(A[1]),cmp);
      	     qsort(B+1,N,sizeof(A[1]),cmp);
        	 qsort(C+1,N,sizeof(A[1]),cmp);
}
int dfs(int i,int j,int k)
{
	if(A[i].pos!=B[j].pos&&A[i].pos!=C[k].pos&&B[j].pos!=C[k].pos)
		return A[i].l+B[j].l+C[k].l;
	if(A[i].pos==B[j].pos)
        return max(dfs(i+1,j,k),dfs(i,j+1,k));
    if(A[i].pos==C[k].pos)
        return max(dfs(i+1,j,k),dfs(i,j,k+1));
    if(C[k].pos==B[j].pos)
        return max(dfs(i,j+1,k),dfs(i,j,k+1));
}
void solve()
{
	int ans=dfs(1,1,1);
	printf("%d\n",ans);
}
int main()
{
	while(cin>>N)
	{
		input();
		solve();
	}
}