ZOJ1425 POJ1692 Crossed Matchings
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很好的dp题。
dp[i][j]表示的是a数组的第i个位置与b数组的第j个位置时最大的交叉匹配数。
/******************************************************************************* # Author : Neo Fung # Email : neosfung@gmail.com # Last modified: 2012-03-22 20:43 # Filename: ZOJ1425 POJ1692 Crossed Matchings.cpp # Description : ******************************************************************************/#ifdef _MSC_VER#define DEBUG#define _CRT_SECURE_NO_DEPRECATE#endif#include <fstream>#include <stdio.h>#include <iostream>#include <string.h>#include <string>#include <limits.h>#include <algorithm>#include <math.h>#include <numeric>#include <functional>#include <ctype.h>#define MAX 1010using namespace std;int array_a[MAX],array_b[MAX];int dp[MAX][MAX];int main(void){#ifdef DEBUG freopen("../stdin.txt","r",stdin); freopen("../stdout.txt","w",stdout); #endif int ncases,n,m;scanf("%d",&ncases);while(ncases--){scanf("%d%d",&n,&m);for(int i=1;i<=n;++i)scanf("%d",&array_a[i]);for(int i=1;i<=m;++i)scanf("%d",&array_b[i]);memset(dp,0,sizeof(dp));for(int i=1;i<=n;++i)for(int j=1;j<=m;++j){dp[i][j]=max(dp[i-1][j],dp[i][j-1]);if(array_a[i]==array_b[j])continue; //找到一个交叉匹配int k=i-1,l=j-1;for(;k>0;--k)if(array_a[k]==array_b[j])break;for(;l>0;--l)if(array_a[i]==array_b[l])break; //如果找到一个交叉匹配if(k>0 && l>0)dp[i][j]=max(dp[i][j],dp[k-1][l-1]+1);}printf("%d\n",dp[n][m]*2);} return 0;}
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