【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

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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

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();}}