Crossed Matchings Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2838   Accepted: 1840 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

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題意：

給出兩行數(shù)，求上下匹配的最多組數(shù)是多少。匹配規(guī)則1.匹配對的數(shù)字必須相同2.每個(gè)匹配必須有且只能有一個(gè)匹配與之相交叉，且相交叉的兩組匹配數(shù)字必須不同3.一個(gè)數(shù)最多只能匹配一次

題解：

一開始我以為是個(gè)二分匹配的題目，后來想了好久不知道怎么處理第二個(gè)條件。

這題其實(shí)是動(dòng)態(tài)規(guī)劃題。分析：用dp[i][j]表示第一行取i個(gè)數(shù)，第二行取j個(gè)數(shù)字的最多匹配項(xiàng)對于某個(gè)dp[i][j]:1.不匹配第一行i個(gè)，或不匹配第二行第j個(gè)：dp[i][j]=Max(dp[i-1][j],dp[i][j-1])2.如果a[i]==b[j],不產(chǎn)生新匹配，匹配結(jié)果為1的值3.若a[i]!=b[j]:a.則第一行從i往前掃，直到掃到第一個(gè)a[k1]==b[j](k1 b.同理，第二行從j往前掃，直到掃到第一個(gè)b[k2]==a[i](k2 若找不到這樣的k1,k2則不能才產(chǎn)生新匹配，跳過若存在這樣的k1,k2,此時(shí)匹配（a[i],b[k2]）、(a[k1],b[j])匹配，才生新的匹配情形,匹配數(shù)量為：dp[k1-1][k2-1]+2。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<vector>#include<queue>#include<stack>using namespace std;#define rep(i,a,n) for (int i=a;i<n;i++)#define per(i,a,n) for (int i=n-1;i>=a;i--)#define pb push_back#define fi first#define se secondtypedef vector<int> VI;typedef long long ll;typedef pair<int,int> PII;const int inf=0x3fffffff;const ll mod=1000000007;const int maxn=100+10;int n,m;int a[maxn],b[maxn];int d[maxn][maxn];int main(){    int cas;    scanf("%d",&cas);    while(cas--)    {        scanf("%d%d",&n,&m);        rep(i,1,n+1) scanf("%d",&a[i]);        rep(i,1,m+1) scanf("%d",&b[i]);        memset(d,0,sizeof(d));        rep(i,2,n+1) rep(j,2,m+1)        {            d[i][j]=max(d[i][j-1],d[i-1][j]);            if(a[i]==b[j]) continue;            else            {                int p1=0,p2=0;                for(int k=i-1;k>0;k--)                {                    if(a[k]==b[j])                    {                        p1=k;                        break;                    }                }                for(int l=j-1;l>0;l--)                {                    if(b[l]==a[i])                    {                        p2=l;                        break;                    }                }                if(p1&&p2) d[i][j]=max(d[i][j],d[p1-1][p2-1]+2);            }        }        printf("%d/n",d[n][m]);    }        return 0;}