题目描述:

1045 Favorite Color Stripe (30 分)  

Eva is trying to make her own color stripe out of a given one. She would like to keep only her favorite colors in her favorite order by cutting off those unwanted pieces and sewing the remaining parts together to form her favorite color stripe.

It is said that a normal human eye can distinguish about less than 200 different colors, so Eva's favorite colors are limited. However the original stripe could be very long, and Eva would like to have the remaining favorite stripe with the maximum length. So she needs your help to find her the best result.

Note that the solution might not be unique, but you only have to tell her the maximum length. For example, given a stripe of colors {2 2 4 1 5 5 6 3 1 1 5 6}. If Eva's favorite colors are given in her favorite order as {2 3 1 5 6}, then she has 4 possible best solutions {2 2 1 1 1 5 6}, {2 2 1 5 5 5 6}, {2 2 1 5 5 6 6}, and {2 2 3 1 1 5 6}.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤) which is the total number of colors involved (and hence the colors are numbered from 1 to N). Then the next line starts with a positive integer M (≤) followed by M Eva's favorite color numbers given in her favorite order. Finally the third line starts with a positive integer L (≤) which is the length of the given stripe, followed by L colors on the stripe. All the numbers in a line a separated by a space.

Output Specification:

For each test case, simply print in a line the maximum length of Eva's favorite stripe.

Sample Input:

6
5 2 3 1 5 6
12 2 2 4 1 5 5 6 3 1 1 5 6

 

Sample Output:

7

思路:要求两个序列的公共部分,明显公共最长子序列变式啊,不过这道题的要求是中间可以穿相同的数

所以v1[i]==v2[j]的时候递推由dp[i][j]=dp[i-1][j-1]+1变为dp[i][j]=dp[i][j-1]就OK啦

AC代码:

#include<iostream>
#include<queue>
#include<string.h>
#include<string>
#include<map>
#include<unordered_map>
#include<vector>
#include<algorithm>
#include<cmath>
#include<set>
using namespace std;
const int maxn = 205;
const int maxm = 100 + 1;
#define INT_MAX 0x7777777
typedef long long ll;
inline int read()
{
    int X = 0; bool flag = 1; char ch = getchar();
    while (ch < '0' || ch>'9') { if (ch == '-') flag = 0; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { X = (X << 1) + (X << 3) + ch - '0'; ch = getchar(); }
    if (flag) return X;
    return ~(X - 1);
}
int v1[205], v2[10005];
int dp[205][10005];
int main() {
    int n = read(),m = read();;
    for (int i = 1; i <= m; i++) {
        v1[i] = read();
    }
    int l = read();
    for (int i = 1; i <= l; i++) {
        v2[i] = read();
    }
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <=l; j++) {
            if (v1[i] == v2[j]) {
                //dp[i][j]=dp[i-1][j-1]+1,这里换回去就完全是最长公共子序列了
                dp[i][j] = dp[i][j - 1] + 1;//满足题目要求的可重复选择
            }
            else {
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
    }
    cout << dp[m][l] << endl;
    return 0;
}