POJ 1692 Crossed Matchings (DP) #by Plato

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http://poj.org/problem?id=1692

题意:两个集合分别有N1N2个元素,求最大匹配交叉对的数目。(匹配交叉对:ab中有两个元素相等,称其为value-match;现在是要有两个va-match,vb-match,并且他们相交,va!=vb

Idea:

类似LCS

f[i] = MAX{f[i-1][j-1],f[i][j-1],f[i-1][j] }

f[i] = MAX{f[i][j],f[m-1][n-1]+1}

(m = preA[i-1][b[j]];n =preB[j-1][a[i]];m != 0 & n != 0& a[i] != b[j])

然后pre怎么算呢?

直接枚举pre的两维然后再确定值,N^3;

观察下,发现其实也有递推关系的。

#include <cstdio>
#include <iostream>
#include <fstream>
#include <cstring>
using namespace std;

#define MAX(x,y) ((x)>(y)?(x):(y))
#define MIN(x,y) ((x)<(y)?(x):(y))

int a[109],b[109];

void calpre(int n,int pre[109][109],int t[109])
{
    memset(pre,sizeof(pre));
    for (int i = 1;i <= n;i++)
    {
        for (int j = 1;j <= 100;j++)
        {
            pre[i][j] = pre[i-1][j];
        }
        pre[i][t[i]] = i;//pre[i][a[i]] = i; 。。。
    }
}

int main()
{
    freopen("test.txt","r",stdin);
    int M,N1,N2;
    scanf("%d",&M);
    while(M--)
    {
        scanf("%d%d",&N1,&N2);
        for (int i = 1;i <= N1;i++)
            scanf("%d",&a[i]);
        for (int i = 1;i <= N2;i++)
            scanf("%d",&b[i]);

        static int preA[109][109],preB[109][109];
        calpre(N1,preA,a);
        calpre(N2,preB,b);

        static int f[109][109];
        int m,n;
        memset(f,sizeof(f));
        for (int i = 1;i <= N1;i++)
        {
            for (int j = 1;j <= N2;j++)
            {
                f[i][j] = MAX(f[i-1][j-1],f[i-1][j]);
                f[i][j] = MAX(f[i][j],f[i][j-1]);
                m = preA[i-1][b[j]];
                n = preB[j-1][a[i]];
                if (m && n && a[i] != b[j])//if (m+n && a[i] != b[j]) 把有一个为0写成了同时为0
                {
                    f[i][j] = MAX(f[i][j],f[m-1][n-1] + 1);//f[i][j] = MAX(f[i][j],f[m][n]+1); m、n都应该减1的
                }
            }
        }

        printf("%d\n",f[N1][N2]*2);
    }

    return 0;
}
/*
1st:WA!方法是对,但是有几个地方写挫了。(1个地方是数学表达,1个代码表达,思路还是不够严密)
2nd:AC!
*/

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