POJ1692 Crossed Matchings（dp）

Crossed Matchings

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 2938 Accepted: 1910

Description

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 input 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 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

36 61 3 1 3 1 33 1 3 1 3 14 41 1 3 3 1 1 3 3 12 111 2 3 3 2 4 1 5 1 3 5 103 1 2 3 2 4 12 1 5 5 3 

Sample Output

608

Source

Tehran 1999

思路：dp[i][j]:=第一行前i个数与第二行前j个数的最大匹配数；

状态转移方程：

1）当a[i]==b[j]时，显然dp[i][j]=max(dp[i][j-1],dp[i-1][j])；

2）当a[i]!=b[j]时，假设a[i]和b[q]相匹配（q<j），b[j]和a[p]相匹配（p<i），则显然dp[i][j]=max{dp[p-1][q-1]+2}

根据状态转移方程观察一下时间复杂度，在 2）中显然需要4个for循环，时间复杂度为O（m^2*n^2），显然时间过大，需要优化。

怎么优化 2）呢？显然我们选择相匹配的肯定是在满足相匹配时最大的位置q和p，我们可以通过预处理找出最大的位置q和p，设定两个表示位置的数组q[maxn][maxn],p[maxn][maxn]，其中q[i][j]表示在数组b中下标小于j且与a[i]相匹配的的数的最大位置，而p[j][i]表示在数组a中下标小于i且与b[j]相匹配的数的最大位置；

经过优化显然时间复杂度为O（m*n）；

AC代码：

#include <bits/stdc++.h>//提交poj把头文件改一下using namespace std;const int maxn=1001;int dp[maxn][maxn],p[maxn][maxn],q[maxn][maxn];int a[maxn],b[maxn];int m,n,T;int main(){    scanf("%d",&T);while (T--) {cin>>m>>n;for(int i=1;i<=m;i++)    cin>>a[i];for(int j=1;j<=n;j++)    cin>>b[j];    for(int i=1;i<=m;i++)    for(int j=1;j<=n;j++)    {    if(a[i]==b[j-1])        q[i][j]=j-1;    else        q[i][j]=q[i][j-1];}for(int j=1;j<=n;j++)    for(int i=1;i<=m;i++)    {    if(b[j]==a[i-1])        p[j][i]=i-1;    else        p[j][i]=p[j][i-1];}        for(int i=1;i<=m;i++)        for(int j=1;j<=n;j++)        {        dp[i][j]=max(dp[i][j-1],dp[i-1][j]);        if(a[i]!=b[j]&&p[j][i]>0&&q[i][j]>0)//显然应该还要满足位置>0    dp[i][j]=max(dp[i][j],dp[p[j][i]-1][q[i][j]-1]+2); }cout<<dp[m][n]<<endl;}return 0;}