Poj 1692 crossed matchings (DP ).

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Question: Give a sequence of two numbers, and find the maximum number of upper and lower matching groups.
Matching rules:
1. The matching numbers must be the same.
2. Each matching must have only one matching that is matched with each other, and the matching numbers must be different.
3. A single number can only be matched once.

For the adaptation of the longest common subsequence, F [I] [J] indicates the optimal value of matching between the first I digit of the first sequence and the first J digit of the second sequence.

State Transfer: F [I] [J] = max (F [P-1] [q-1] + 2, max (F [I-1] [J], f [I] [J-1])

& (A [p] = B [J], B [Q] = A [I]);

* Updating f [I] [J] with f [P-1] [q-1] ensures that a single number can only be matched once.

Question link: http://poj.org/problem? Id = 1692

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#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#define N 1000using namespace std;int f[N][N];int a[N],b[N];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        int i,j,p,q;        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        for(j=1;j<=m;j++)            scanf("%d",&b[j]);        memset(f,0,sizeof(f));        for(i=1;i<=n;i++)        {            for(j=1;j<=m;j++)            {                for(q=j-1;q>0;q--)                    if(b[q]==a[i]) break;                for(p=i-1;p>0;p--)                    if(a[p]==b[j]) break;                if(p>0 && q>0 && a[i]!=b[j])   //~~~                    f[i][j]=max(f[p-1][q-1]+2,max(f[i][j-1],f[i-1][j]));                else                    f[i][j]=max(f[i][j-1],f[i-1][j]);            }        }        printf("%d\n",f[n][m]);    }    return 0;}