POJ 4044 寻找最长连续公共子序列

         题意很简单，就是对两组数据排序后寻找最长连续公共子序列，排序前应该先去重。由于数据范围很小（30），所以可以完全暴力求解。随后将最长连续公共子序列的各个数按各位升序排序即可。

ac代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>

using namespace std;

const int N = 35;
int numa[N],numb[N];

bool cmp(int a,int b){
	return a>b;
}

bool cmp2(int a,int b){
	if(a%10 == b%10)
		return a<b;
	return a%10 < b%10;
}

int main(){
	//freopen("1.txt","r",stdin);
	int numcase,n,m;
	scanf("%d",&numcase);
	for(int k = 1;k <= numcase;++k){
	   memset(numa,-1,sizeof(numa));
	   memset(numb,-1,sizeof(numb));
	   scanf("%d%d",&n,&m);
	   for(int i = 0;i < n;++i){
	     scanf("%d",&numa[i]);
		 for(int j = 0;j < i;++j){
			 if(numa[j] == numa[i]){
			   i--;
			   n--;
			 }
		 }
	   }
	   for(int i = 0;i < m;++i){
	     scanf("%d",&numb[i]);
		 for(int j = 0;j < i;++j){
			 if(numb[j] == numb[i]){
			   i--;
			   m--;
			 }
		 }
	   }
	   sort(numa,numa+n,cmp);
	   sort(numb,numb+m,cmp);
	   int mmax = 0,abegin=0,aend=0;
	   for(int i = 0;i < n;++i){
		   for(int j = 0;j < m;++j){
			   if(numa[i] == numb[j]){
     		      int k = 0;
				  while(i+k < n && j+k < m && numa[i+k] == numb[j+k]){
				    k++;
				  }
				  if(k > mmax){
				    mmax = k;
					abegin = i;
					aend = i + k;
				  }
			   }
		   }
	   }
	   if(mmax == 0)puts("NONE");
	   else{
	     for(int i = abegin;i < aend;++i)
			 printf("%d ",numa[i]);
		 printf("\n");
		 sort(numa+abegin,numa+aend,cmp2);
		 for(int i = abegin;i < aend;++i)
			 printf("%d ",numa[i]);
		 printf("\n");
	   }
	}
	return 0;
}