zoj 1425 最大交叉匹配

Crossed MatchingsTime Limit: 2 Seconds Memory Limit: 65536 KB
There are two rows of positive integer numbers. We can draw one line segment between any two equal numbers, with values r, if one of them is located in the first row and the other one is located in the second row. We call this line segment an r-matching segment. The following figure shows a 3-matching and a 2-matching segment.



We want to find the maximum number of matching segments possible to draw for the given input, such that:

1. Each a-matching segment should cross
exactly one b-matching segment, where a != b.

2. No two matching segments
can be drawn from a number. For example, the following matchings are not
allowed.



Write a program to compute the maximum number of matching segments for the
input data. Note that this number is always even.

Input

The first line of the file is the number M, which is
the number of test cases (1 <= M <= 10). Each test case has three lines.
The first line contains N1 and N2, the number of integers on the first and the
second row respectively. The next line contains N1 integers which are the
numbers on the first row. The third line contains N2 integers which are the
numbers on the second row. All numbers are positive integers less than 100.

Output

Output file should have one separate line for each
test case. The maximum number of matching segments for each test case should be
written in one separate line.

Sample Input

3
6 6
1 3 1 3 1 3
3 1 3 1 3 1
4
4
1 1 3 3
1 1 3 3
12 11
1 2 3 3 2 4 1 5 1 3 5 10
3 1 2 3 2 4 12
1 5 5 3

Sample Output

6
0
8

/*
zoj 1425 最大交叉匹配
题目大意：给两行序列，求他们的最大交叉数*2。
定义交叉：上下相同的数字连线，两条这样的线相交，
每个数字只能用与一条线，且两条线的数字不等。
定义dp(i,j)表示第一行至i，第二行至j，的最大交叉数，
dp(i,j)=max(dp(i-1,j-1)((i,j都不是))dp(i-1,j)(i不是),dp(i,j-1)(j不是),dp(n-1,m-1)+1(n-j,m-i交叉))。
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn=105;
int dp[maxn][maxn];
int a[maxn],b[maxn];

inline int max(int a,int b){return a>b?a:b;}

int main()
{
int t,i,j,k,l,n,m;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(i=1;i<=n;i++) scanf("%d",a+i);
for(i=1;i<=m;i++) scanf("%d",b+i);
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
dp[i][j]=max(dp[i-1][j-1],max(dp[i-1][j],dp[i][j-1]));
if(a[i]==b[j]) continue;//两条交叉线段必须数字不同
k=i-1;l=j-1;
while(k>0 && a[k]!=b[j]) k--;
while(l>0 && a[i]!=b[l]) l--;
if(k && l) dp[i][j]=max(dp[k-1][l-1]+1,dp[i][j]);
}
}
printf("%d\n",dp
[m]*2);
}
return 0;
}